Please help!!! I’ll mark you as brainliest if correct
2 answers:
You might be interested in
Answer:
P-value (t=2.58) = 0.0066.
Note: as we are using the sample standard deviation, a t-statistic is appropiate instead os a z-statistic.
As the P-value (0.0066) is smaller than the significance level (0.05), the effect is significant.
The null hypothesis is rejected.
There is enough evidence to support the claim that the population mean μ exceeds 2.4.
Step-by-step explanation:
This is a hypothesis test for the population mean.
The claim is that the population mean μ exceeds 2.4.
Then, the null and alternative hypothesis are:
The significance level is 0.05.
The sample has a size n=45.
The sample mean is M=2.5.
As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=0.26.
The estimated standard error of the mean is computed using the formula:
Then, we can calculate the t-statistic as:
The degrees of freedom for this sample size are:
This test is a right-tailed test, with 44 degrees of freedom and t=2.58, so the P-value for this test is calculated as (using a t-table):
As the P-value (0.0066) is smaller than the significance level (0.05), the effect is significant.
The null hypothesis is rejected.
There is enough evidence to support the claim that the population mean μ exceeds 2.4.