1/2a - 6 = 3 + a solve for a plssss
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Answer:
The null hypothesis failed to be rejected.
There is not enough evidence to support the claim that the mean daily conmute time for males is significantly bigger than the mean daily conmute time for females (P-value=0.155).
Step-by-step explanation:
The data is:
Males n1=207 M1=31.6 s1=24.0
Females n2=253 M2=29.3 s2=24.3
This is a hypothesis test for the difference between populations means.
The claim is that the mean daily conmute time for males is significantly bigger than the mean daily conmute time for females.
Then, the null and alternative hypothesis are:
The significance level is 0.05.
The sample 1 (males), of size n1=207 has a mean of 31.6 and a standard deviation of 24.
The sample 2 (females), of size n2=253 has a mean of 29.3 and a standard deviation of 24.3.
The difference between sample means is Md=2.3.
The estimated standard error of the difference between means is computed using the formula:
Then, we can calculate the t-statistic as:
The degrees of freedom for this test are:
This test is a right-tailed test, with 458 degrees of freedom and t=1.017, so the P-value for this test is calculated as (using a t-table):
As the P-value (0.155) is bigger than the significance level (0.05), the effect is not significant.
The null hypothesis failed to be rejected.
There is not enough evidence to support the claim that the mean daily conmute time for males is significantly bigger than the mean daily conmute time for females.