Graded Assignment 11

Question 1

Problem: The average marks scored by students… is reported to be 400 with a standard deviation of 5. You suspect the average may be lower, possibly 390… What sample size do you need for a test at significance level 0.05 and power 0.95?

Concept: Sample Size for Hypothesis Testing (Mean) To calculate the sample size ($n$) required to detect a specific difference between the null hypothesis mean ($\mu_0$) and an alternative mean ($\mu_1$) with a specific Power ($1-\beta$) and Significance Level ($\alpha$), we use the following formula for a one-sided Z-test:

• $\sigma$: Population standard deviation.
• $z_{\alpha}$: Critical Z-value for the significance level (Type I error).
• $z_{\beta}$: Critical Z-value for the power (Type II error).
• $\mu_0 - \mu_1$: The “Effect Size” or difference we want to detect.

Solution:

1. Identify parameters:

   • $\mu_0 = 400$
   • $\mu_1 = 390$
   • $\sigma = 5$
   • $\alpha = 0.05$ (Significance) $\rightarrow$ From Z-table, $z_{0.05} = 1.645$ (for one-tail).
   • Power = $0.95 \rightarrow \beta = 1 - 0.95 = 0.05 \rightarrow z_{0.05} = 1.645$.

2. Apply Formula:

   $$n = \left( \frac{5 (1.645 + 1.645)}{400 - 390} \right)^2$$$$n = \left( \frac{5 (3.29)}{10} \right)^2$$$$n = \left( \frac{16.45}{10} \right)^2$$$$n = (1.645)^2 \approx 2.706$$

3. Round Up: Sample sizes must be integers, and we always round up to ensure the power requirement is met.

Answer: 3

Question 2

Problem: Find the critical value $c$. Enter the answer correct to two decimal places.

Concept: Critical Value ($c$) in Sample Mean Distribution The critical value $c$ is the specific sample mean ($\bar{x}$) that serves as the boundary between rejecting and failing to reject the null hypothesis. For a lower-tailed test (suspecting average is lower), the critical region is on the left.

Solution:

1. Identify parameters:

   • $\mu_0 = 400$
   • $z_{\alpha} = 1.645$ (for $\alpha=0.05$)
   • $\sigma = 5$
   • $n = 3$ (Calculated in Question 1).

2. Calculate Standard Error (SE):

   $$SE = \frac{\sigma}{\sqrt{n}} = \frac{5}{\sqrt{3}} \approx 2.8868$$

3. Calculate $c$:

   $$c = 400 - (1.645 \times 2.8868)$$$$c = 400 - 4.748$$$$c = 395.252$$

Answer: 395.25

Question 3

Problem: The proportion of adults… is estimated to be $p=0.6$. Test against alternative $p < 0.6$… What sample size is needed for a test (against alternative $p=0.4$) at significance level 0.10 and power 0.90?

Concept: Sample Size for Proportions When dealing with proportions, the variance depends on $p$. The formula for sample size includes variances under both the null ($p_0$) and alternative ($p_1$) hypotheses.

Solution:

1. Identify parameters:

   • $p_0 = 0.6 \Rightarrow q_0 = 0.4$
   • $p_1 = 0.4 \Rightarrow q_1 = 0.6$
   • $\alpha = 0.10 \Rightarrow z_{0.10} = 1.282$ (One-tail)
   • Power = $0.90 \Rightarrow \beta = 0.10 \Rightarrow z_{0.10} = 1.282$

2. Calculate Terms:

   • Numerator Term 1: $1.282\sqrt{0.6 \times 0.4} = 1.282\sqrt{0.24} \approx 1.282(0.4899) \approx 0.628$
   • Numerator Term 2: $1.282\sqrt{0.4 \times 0.6} = 1.282\sqrt{0.24} \approx 0.628$
   • Denominator: $0.6 - 0.4 = 0.2$

3. Apply Formula:

   $$n = \left( \frac{0.628 + 0.628}{0.2} \right)^2$$$$n = \left( \frac{1.256}{0.2} \right)^2$$$$n = (6.28)^2 \approx 39.43$$

4. Round Up: $n = 40$

Answer: 40

Question 4

Problem: Find the critical value at a significance level of 0.10. Enter the answer correct to two decimal places.

Concept: Critical Value for Proportion We need to find the specific proportion value ($p_c$) below which we reject the Null Hypothesis.

Solution:

1. Parameters:

   • $p_0 = 0.6$
   • $z_{0.10} = 1.282$
   • $n = 40$ (from Question 3)

2. Calculate:

   $$c = 0.6 - 1.282 \sqrt{\frac{0.6 \times 0.4}{40}}$$$$c = 0.6 - 1.282 \sqrt{0.006}$$$$c = 0.6 - 1.282 (0.07746)$$$$c = 0.6 - 0.0993$$$$c = 0.5007$$

Answer: 0.50

Question 5

Problem: $X \sim \text{Normal}(\mu, 9)$. $n=100$, sample mean $\bar{x} = 11.8$. $H_0: \mu = 10.5$ vs $H_1: \mu \neq 10.5$. Choose correct options.

Concept: Z-Test for Mean (Two-Tailed)

• Variance: Given as 9, so $\sigma = \sqrt{9} = 3$.
• Test Statistic: Since population variance is known, we use the Z-test.
• Direction: The alternative hypothesis has $\neq$, making it a Two-tailed test.

Calculation:

Decision:

• A Z-score of 4.33 is extreme.
• Critical Z for $\alpha=0.05$ is $\pm 1.96$.
• Critical Z for $\alpha=0.10$ is $\pm 1.645$.
• Since $4.33 > 1.96$, we Reject $H_0$ at both levels.

Correct Options:

• Test statistic used is sample mean, $\bar{X}$.
• Two tailed z-test is used.
• Reject $H_0$ at a significance level of 0.10.
• Reject $H_0$ at a significance level of 0.05.

Question 6

Problem: Sample of 36 marshmallow packets, average 145g, SD 5g. Test $H_0: \mu = 150$ vs $H_1: \mu < 150$ at 0.05 level.

Solution:

1. Calculate Z-score: $$Z = \frac{145 - 150}{5 / \sqrt{36}} = \frac{-5}{5/6} = -6$$
2. Critical Value: For left-tail $\alpha=0.05$, $z_{crit} = -1.645$.
3. Conclusion: $-6 < -1.645$. The result falls deep in the rejection region. We Reject the Null Hypothesis. This means we accept the Alternative Hypothesis: The average weight is less than 150 grams.

Answer Option:

• On average, it weighs less than 150 grams.

Question 7

Problem: Survey of 225 students, 89.4% participated in extracurriculars. Can we conclude at 1% level of significance that 90% of the students have participated?

Concept: Interpretation of “Concluding”

• $H_0: p = 0.90$
• $H_1: p \neq 0.90$ (Implied two-sided unless “less than” is specified, but let’s check the Z score).
• $\hat{p} = 0.894$.

Calculation:

Conclusion:

• $|Z| = 0.3$. The critical value for 1% ($\alpha=0.01$) is $2.576$.
• Since $0.3 < 2.576$, we Fail to Reject $H_0$.
• Crucial Logic: In statistics, we never “Conclude that $H_0$ is true.” We only conclude that we don’t have enough evidence to say it’s false.
• However, if the question asks “Can we conclude that 90% have participated?”, it is asking if we can statistically affirm the Null Hypothesis as a proven fact. The answer to “Can we conclude [Null]?” is almost always No.
• Alternatively, if the question implies “Is the claim of 90% plausible?”, the answer is Yes.
• Standard Academic Answer: “No”. (Because we cannot prove the Null Hypothesis, we can only fail to disprove it).

Answer: No

Question 8

Problem: Test if average lifetime exceeds 1000 hours ($H_1: \mu > 1000$). Calculate the P-value.

• Note: The options suggest a calculated Z-score of 2. ($F_Z(2)$ appears in the options).
• Let’s check: $1 - F_Z(2) = 1 - 0.9772 = 0.0228$.

Concept: P-Value for Right-Tailed Test For a test checking if the mean is greater than a value (Right/Upper Tail), the P-value is the area under the curve to the right of the calculated Z-score.

Solution: Looking at the options, the one that represents “Area to the right of Z=2” is:

This equals $1 - 0.9772 = 0.0228$.

Answer Options:

• $1 - F_Z(2)$
• $0.0228$ (Both represent the same correct value, but “0.0228” is the calculated final answer).

Question 10 (Blank 1)

Prompt: Since the population standard deviation is known, the sampling distribution of $\bar{X}$ under the null hypothesis will be [1]…

Concept: Sampling Distribution properties When the parent population is normally distributed, the sampling distribution of the sample mean ($\bar{X}$) is also normally distributed, regardless of the sample size. This is a fundamental property of the normal distribution.

Answer: 2 (Option 2 corresponds to “Normal”)

Question 11 (Blank 2)

Prompt: …distribution with mean [2]…

Concept: Mean of the Sampling Distribution ($\mu_{\bar{x}}$) The mean of the sampling distribution is always equal to the population mean ($\mu$).

Given in the problem, $\mu = 100$.

Answer: 4 (Option 4 corresponds to “100”)

Question 12 (Blank 3)

Prompt: …and variance [3].

Concept: Variance of the Sampling Distribution ($\sigma^2_{\bar{x}}$) The variance of the sample mean is the population variance divided by the sample size ($n$).

• Population Standard Deviation ($\sigma$) = 2
• Population Variance ($\sigma^2$) = $2^2 = 4$
• Sample size ($n$) = 9

Answer: 6 (Option 6 corresponds to “4/9”)

Question 13 (Blank 4)

Prompt: $\alpha = P(\bar{X} > \text{[4]} \text{ or } …)$

Concept: Rejection Region (Upper Tail) The problem states that we accept $H_0$ if the sample mean is between 98.5 and 101.5. Conversely, we reject $H_0$ (which defines $\alpha$, the significance level) if the mean is outside this range. The “greater than” condition corresponds to the upper boundary of the acceptance interval.

• Upper boundary = 101.5

Answer: 8 (Option 8 corresponds to “101.5”)

Question 14 (Blank 5)

Prompt: … or $\bar{X} < \text{[5]} | \mu = 100)$

Concept: Rejection Region (Lower Tail) Similar to the previous step, we reject $H_0$ if the sample mean is less than the lower boundary of the acceptance interval.

• Lower boundary = 98.5

Answer: 7 (Option 7 corresponds to “98.5”)

Question 15 (Blank 6)

Prompt: $= P( | \text{[6]} | > \dots )$

Concept: Standardization (Z-score) To calculate probabilities, we convert the sample mean ($\bar{X}$) into a standard Z-score. The formula for Z when dealing with a sample mean is:

Substituting the values:

• $\mu = 100$
• Standard Error ($\sigma / \sqrt{n}$) = $\sqrt{4/9} = 2/3$

So, the expression for Z is:

Answer: 10 (Option 10 corresponds to $\frac{\bar{X} - 100}{2/3}$)

Question 16 (Blank 7)

Prompt: $= P( | Z | > \text{[7]} )$

Concept: Calculating Critical Z-value We need to find the Z-score corresponding to our critical boundaries (98.5 and 101.5). Using the upper boundary (101.5):

So, we are looking for the probability that the absolute value of Z is greater than 2.25.

Answer: 13 (Option 13 corresponds to “2.25”)

Question 17 (Blank 8)

Prompt: $= 1 - P( \text{[8]} < Z < \dots )$

Concept: Complement Rule The probability of being in the rejection region ($|Z| > 2.25$) is equal to 1 minus the probability of being in the acceptance region. The acceptance region is the interval between the negative and positive critical Z-values.

• Lower limit Z = -2.25

Answer: 12 (Option 12 corresponds to “-2.25”)

Question 18 (Blank 9)

Prompt: $= 1 - P( \dots < Z < \text{[9]} )$

Concept: Acceptance Region Upper Limit Continuing from the previous step, the upper limit of the acceptance region in terms of Z-score is the positive critical value.

• Upper limit Z = +2.25

Answer: 13 (Option 13 corresponds to “2.25”)

Question 19 (Blank 10)

Prompt: Calculate the value for [10] (Enter the answer correct to two decimal places.)

Concept: Final Calculation of Significance Level ($\alpha$) We need to evaluate: $1 - P(-2.25 < Z < 2.25)$.

1. Find the area under the normal curve between -2.25 and 2.25.
   • Using a Z-table, $P(Z < 2.25) \approx 0.9878$.
   • $P(Z < -2.25) \approx 0.0122$.
   • Area between = $0.9878 - 0.0122 = 0.9756$.
2. Subtract from 1.
   • $\alpha = 1 - 0.9756 = 0.0244$.

Rounding to two decimal places: 0.02

Answer: 0.02

Question 10 (Blank 1) Concept: Sampling Distribution of the Mean When the population follows a Normal distribution, the sampling distribution of the sample mean ($\bar{X}$) also follows a Normal distribution, regardless of sample size.

Answer: 2 (Option: “Normal”)

Question 11 (Blank 2) Concept: Mean of Sampling Distribution The mean of the sampling distribution ($\mu_{\bar{x}}$) is equal to the population mean ($\mu$).

Answer: 4 (Option: “100”)

Question 12 (Blank 3) Concept: Variance of Sampling Distribution The variance of the sample mean is the population variance divided by the sample size ($n$).

• Population Variance ($\sigma^2$) = $2^2 = 4$
• Sample Size ($n$) = 9
• Variance ($\sigma_{\bar{x}}^2$) = $4/9$

Answer: 6 (Option: “4/9”)

Question 13 (Blank 4) Concept: Rejection Region (Upper Bound) We reject the null hypothesis if the sample mean falls outside the acceptance interval $[98.5, 101.5]$. The upper rejection region is when $\bar{X}$ is greater than the upper limit.

Answer: 8 (Option: “101.5”)

Question 14 (Blank 5) Concept: Rejection Region (Lower Bound) Similarly, the lower rejection region is when $\bar{X}$ is less than the lower limit of the acceptance interval.

Answer: 7 (Option: “98.5”)

Question 15 (Blank 6) Concept: Z-Score Formula To standardize the sample mean, we use the Z-score formula:

The question asks for the absolute value term inside the probability function $P(|\dots| > \dots)$, which represents the standardized distance from the mean.

Answer: 10 (Option: $\frac{\bar{X} - 100}{2/3}$)

Question 16 (Blank 7) Concept: Critical Z-Value Calculate the Z-score for the boundary value (101.5).

So, we reject if $|Z| > 2.25$.

Answer: 13 (Option: “2.25”)

Question 17 (Blank 8) Concept: Acceptance Region (Lower Z) The probability of rejection is $1 - P(\text{Acceptance})$. The acceptance region corresponds to Z-scores between the negative and positive critical values. Lower Z boundary = $-2.25$

Answer: 12 (Option: “-2.25”)

Question 18 (Blank 9) Concept: Acceptance Region (Upper Z) Upper Z boundary = $+2.25$

Answer: 13 (Option: “2.25”)

Question 19 (Blank 10) Concept: Calculating Significance Level ($\alpha$)

Using a standard normal table:

• $P(Z < 2.25) \approx 0.9878$
• $P(Z < -2.25) \approx 0.0122$
• Area between = $0.9878 - 0.0122 = 0.9756$
• $\alpha = 1 - 0.9756 = 0.0244$ Rounding to two decimal places:

Answer: 0.02

Question 21 (Blank 1) Concept: Rejection Criteria (Upper) The rejection rule does not change. We still reject if the sample mean is greater than the upper critical limit established in the design of the test.

Answer: 3 (Option: “101.5”)

Question 22 (Blank 2) Concept: Rejection Criteria (Lower) We also reject if the sample mean is less than the lower critical limit.

Answer: 1 (Option: “98.5”)

Question 23 (Blank 3) Concept: Standardizing Upper Bound with New Mean We calculate the Z-score for $X = 101.5$ using the new mean $\mu = 103$.

So the condition $\bar{X} > 101.5$ becomes $Z > -2.25$.

Answer: 12 (Option: “-2.25”)

Question 24 (Blank 4) Concept: Standardizing Lower Bound with New Mean We calculate the Z-score for $X = 98.5$ using $\mu = 103$.

So the condition $\bar{X} < 98.5$ becomes $Z < -6.75$.

Answer: 7 (Option: “-6.75”)

Question 25 (Blank 5) Concept: Probability Calculation (Term 1) We need to find $P(Z > -2.25)$. Using symmetry, this is equal to $P(Z < 2.25)$. From the Z-table: $P(Z < 2.25) \approx 0.9878$. Rounding to match options:

Answer: 10 (Option: “0.99”)

Question 26 (Blank 6) Concept: Probability Calculation (Term 2) We need to find $P(Z < -6.75)$. A Z-score of -6.75 is extremely far in the left tail. The area to the left is effectively zero.

Answer: 11 (Option: “0”)

Question 27 (Blank 7) Concept: Final Power Calculation Total Power = (Probability from Term 1) + (Probability from Term 2)

This means there is a 99% chance the test will correctly detect that the mean has shifted to 103 minutes.

Answer: 0.99