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Answer:
Critical value: T = 1.337
The 80% confidence interval for the mean repair cost for the stereos is between $45.991 and $86.689
Step-by-step explanation:
We have the standard deviation for the sample. So we use the t-distribution to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 17 - 1 = 16
80% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 16 degrees of freedom(y-axis) and a confidence level of . So we have T = 1.337, which is the critical value
The margin of error is:
M = T*s = 1.337*15.22 = 20.349
In which s is the standard deviation of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 66.34 - 20.349 = $45.991
The upper end of the interval is the sample mean added to M. So it is 66.34 + 20.349 = $86.689
The 80% confidence interval for the mean repair cost for the stereos is between $45.991 and $86.689
Answer:
The degrees of freedom are
And the p value is given by:
Since the p value for this cae is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true means for this case are significantly different
Step-by-step explanation:
When we have two independent samples from two normal distributions with equal variances we are assuming that
And the statistic is given by this formula:
Where t follows a t distribution with degrees of freedom and the pooled variance
is given by this formula:
The system of hypothesis on this case are:
Null hypothesis:
Alternative hypothesis:
We have the following data given:
represent the sample size for group 1
represent the sample size for group 2
represent the sample mean for the group 1
represent the sample mean for the group 2
represent the sample standard deviation for group 1
represent the sample standard deviation for group 2
First we can begin finding the pooled variance:
And the deviation would be just the square root of the variance:
The statistic is givne by:
The degrees of freedom are
And the p value is given by:
Since the p value for this cae is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true means for this case are significantly different