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Math  /  Data & Statistics

QuestionC9 Q5 V3: A random sample of 100 students is taken from a larger population of students in a multi-lecture course and the sample mean (average) of their final grades is 66. Suppose it is known that the population standard deviation is 7. You test whether there is significant evidence that the population average differs from 67 , using a level of significance of 4%4 \%. Choose the most correct statement. Note: You will have to calculate the test statistic by hand, but you can use R to get your pvalue. a. Your pvalue statement is 2P(Z>1.43)2 \mathrm{P}(\mathrm{Z}>1.43), your pvalue is 0.1531 , and your decision is to reject Ho. b. Your pvalue statement is 2P(Z>1.43)2 P(Z>1.43), your pvalue is 0.1531 and your decision is to fail to reject Ho c. Your pvalue statement is 2P(Z>1.43)2 P(Z>1.43), your pvalue is 0.3062 and your decision is to fail to reject Ho. d. Your pvalue statement is 2P(Z>1.43)2 \mathrm{P}(\mathrm{Z}>1.43), your pvalue is 0.0766 and your decision is reject Ho.

Studdy Solution

STEP 1

1. The sample size is n=100 n = 100 .
2. The sample mean is xˉ=66 \bar{x} = 66 .
3. The population standard deviation is σ=7 \sigma = 7 .
4. The null hypothesis H0 H_0 is that the population mean μ=67 \mu = 67 .
5. The alternative hypothesis Ha H_a is that the population mean μ≠67 \mu \neq 67 .
6. The level of significance is α=0.04 \alpha = 0.04 .

STEP 2

1. Calculate the test statistic.
2. Determine the p-value using the test statistic.
3. Make a decision based on the p-value and level of significance.

STEP 3

Calculate the test statistic using the formula:
Z=xˉ−μσnZ = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}
Substitute the given values:
Z=66−677100=−10.7≈−1.43Z = \frac{66 - 67}{\frac{7}{\sqrt{100}}} = \frac{-1}{0.7} \approx -1.43

STEP 4

The test is two-tailed, so the p-value is calculated as:
p-value=2×P(Z>∣1.43∣)\text{p-value} = 2 \times P(Z > |1.43|)
Using statistical software or a Z-table, find P(Z>1.43) P(Z > 1.43) .
Assume P(Z>1.43)≈0.0766 P(Z > 1.43) \approx 0.0766 .
Thus, the p-value is:
2×0.0766=0.15322 \times 0.0766 = 0.1532

STEP 5

Compare the p-value to the level of significance α=0.04 \alpha = 0.04 .
Since 0.1532>0.04 0.1532 > 0.04 , we fail to reject the null hypothesis H0 H_0 .
The most correct statement is:
b. Your pvalue statement is 2P(Z>1.43)2 P(Z>1.43), your pvalue is 0.1531 and your decision is to fail to reject Ho.

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