Question

Use technology and a t-test to test claim about the population mean mu at the given level of significance alpha using the given sample statistics. Assume the population is normally distributed. Claim mu > 76 alpha = 0.01 Sample statistics x = 77.5, s= 3.3, n=29 What are the null and alternative hypotheses? Choose the correct answer below What is te value of the standardized test statistic? The standardized test statistic is (Round to two decimal places as needed) What is the P-value of the test statistic? P-value = (round to three decimal places as needed.) What is the value of the standardized test statistic? The standardized test statistic is . (Round to two decimal places as needed.) What is the P-value of the test statistic? P-value = (Round to three decimal places as needed.) Decide whether to reject or fail to reject the null hypothesis. Choose the correct answer below. Reject H0. There is enough evidence to support the claim. Fail to reject H0. There is not enough evidence to support the claim. Fail to reject H0 There is enough evidence to support the claim. Fail to reject H0 there is not enough evidence to support the claim.

Answer 1

Answer:

a) Null hypothesis:$\mu \leq 76$  

Alternative hypothesis:$\mu > 76$

b) $t=\frac{77.5-76}{\frac{3.3}{\sqrt{29}}}=2.45$  

c) $p_v =P(t_{(28)}>2.45)=0.0104$  

d)  Fail to reject H0. There is not enough evidence to support the claim.

Step-by-step explanation:

Data given and notation  

$\bar X=77.5$ represent the sample mean  

$s=3.3$ represent the sample standard deviation  

$n=29$ sample size  

$\mu_o =76$ represent the value that we want to test  

$\alpha=0.01$ represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

a) State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the population mean is higher than 76, the system of hypothesis are :  

Null hypothesis:$\mu \leq 76$  

Alternative hypothesis:$\mu > 76$  

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

b) Calculate the statistic  

We can replace in formula (1) the info given like this:  

$t=\frac{77.5-76}{\frac{3.3}{\sqrt{29}}}=2.45$  

c) P-value  

We need to calculate the degrees of freedom first given by:

$df=n-1=29-1=28$

Since is a one-side right tailed test the p value would given by:  

$p_v =P(t_{(28)}>2.45)=0.0104$  

We can use the following code in excel to find the answer: "=1-T.DIST(2.45,28,TRUE)"

d) Conclusion  

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

The best option on this case:

Fail to reject H0. There is not enough evidence to support the claim.