How do I solve this problem?
1 answer:
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I hope this helps! It’s been a while since I’ve done slope-intercept though, so I might be a little rusty
Answer:
a. P-value = 0.039.
The null hypothesis is rejected.
At a significance level of 0.05, there is enough evidence to support the claim that the population mean significantly differs from 100.
b. P-value = 0.013.
The null hypothesis is rejected.
At a significance level of 0.05, there is enough evidence to support the claim that the population mean significantly differs from 100.
c. P-value = 0.130.
The null hypothesis failed to be rejected.
At a significance level of 0.05, there is not enough evidence to support the claim that the population mean significantly differs from 100.
Step-by-step explanation:
This is a hypothesis test for the population mean.
The claim is that the population mean significantly differs from 100.
Then, the null and alternative hypothesis are:
The significance level is 0.05.
The sample has a size n=65.
The degrees of freedom for this sample size are:
a. The sample mean is M=103.
As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=11.5.
The estimated standard error of the mean is computed using the formula:
Then, we can calculate the t-statistic as:
This test is a two-tailed test, with 64 degrees of freedom and t=2.103, so the P-value for this test is calculated as (using a t-table):
As the P-value (0.039) is smaller than the significance level (0.05), the effect is significant.
The null hypothesis is rejected.
At a significance level of 0.05, there is enough evidence to support the claim that the population mean significantly differs from 100.
b. The sample mean is M=96.5.
As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=11.
The estimated standard error of the mean is computed using the formula:
Then, we can calculate the t-statistic as:
This test is a two-tailed test, with 64 degrees of freedom and t=-2.565, so the P-value for this test is calculated as (using a t-table):
As the P-value (0.013) is smaller than the significance level (0.05), the effect is significant.
The null hypothesis is rejected.
At a significance level of 0.05, there is enough evidence to support the claim that the population mean significantly differs from 100.
c. The sample mean is M=102.
As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=10.5.
The estimated standard error of the mean is computed using the formula:
Then, we can calculate the t-statistic as:
This test is a two-tailed test, with 64 degrees of freedom and t=1.536, so the P-value for this test is calculated as (using a t-table):
As the P-value (0.13) is bigger than the significance level (0.05), the effect is not significant.
The null hypothesis failed to be rejected.
At a significance level of 0.05, there is not enough evidence to support the claim that the population mean significantly differs from 100.
