Math  /  Data & Statistics

Question10:4/PM Fri Nov 22
You wish to test the following claim ( HaH_{a} ) at a significance level of α=0.01\alpha=0.01. Ho:μ=83.5Ha:μ83.5\begin{array}{l} H_{o}: \mu=83.5 \\ H_{a}: \mu \neq 83.5 \end{array}
You believe the population is normally distributed, but you do not know the standard deviation. You obtain a sample of size n=556n=556 with mean xˉ=85.4\bar{x}=85.4 and a standard deviation of s=16.7s=16.7.
What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic = \square
What is the pp-value for this sample? (Report answer accurate to four decimal places.) p -value = \square The pp-value is... less than (or equal to) α\alpha greater than α\alpha
This test statistic leads to a decision to... reject the null accept the null fail to reject the null
As such, the final conclusion is that... There is sufficient evidence to warrant rejection of the claim that the population mean is not equal to 83.5 . There is not sufficient evidence to warrant rejection of the claim that the population mean is not equal to 83.5 . The sample data support the claim that the population mean is not equal to 83.5. There is not sufficient sample evidence to support the claim that the population mean is not equal to 83.5 . Check Answer

Studdy Solution

STEP 1

1. The population is normally distributed.
2. The sample size is n=556 n = 556 .
3. The sample mean is xˉ=85.4 \bar{x} = 85.4 .
4. The sample standard deviation is s=16.7 s = 16.7 .
5. The null hypothesis is H0:μ=83.5 H_0: \mu = 83.5 .
6. The alternative hypothesis is Ha:μ83.5 H_a: \mu \neq 83.5 .
7. The significance level is α=0.01 \alpha = 0.01 .

STEP 2

1. Calculate the test statistic.
2. Determine the p-value.
3. Compare the p-value with the significance level.
4. Make a decision based on the p-value.
5. State the final conclusion.

STEP 3

Calculate the test statistic using the formula for the t-test when the population standard deviation is unknown:
t=xˉμs/nt = \frac{\bar{x} - \mu}{s / \sqrt{n}}
Substitute the given values:
t=85.483.516.7/556t = \frac{85.4 - 83.5}{16.7 / \sqrt{556}}
Calculate the standard error:
SE=16.75560.707SE = \frac{16.7}{\sqrt{556}} \approx 0.707
Calculate the test statistic:
t1.90.7072.686t \approx \frac{1.9}{0.707} \approx 2.686

STEP 4

Determine the p-value for the calculated test statistic. Since this is a two-tailed test, we need to find the probability that t t is greater than 2.686 or less than -2.686.
Using a t-distribution table or calculator with n1=555 n-1 = 555 degrees of freedom, find the p-value:
p-value0.0073p\text{-value} \approx 0.0073

STEP 5

Compare the p-value with the significance level α=0.01 \alpha = 0.01 :
Since 0.0073<0.01 0.0073 < 0.01 , the p-value is less than α \alpha .

STEP 6

Based on the p-value comparison:
The test statistic leads to a decision to reject the null hypothesis.

STEP 7

State the final conclusion:
There is sufficient evidence to warrant rejection of the claim that the population mean is equal to 83.5.
The test statistic is approximately 2.686 \boxed{2.686} .
The p-value is approximately 0.0073 \boxed{0.0073} .
The p-value is less than (or equal to) α \alpha .
This test statistic leads to a decision to reject the null.
As such, the final conclusion is that there is sufficient evidence to warrant rejection of the claim that the population mean is equal to 83.5.

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