Susan's math teacher assigned x homework problems on Monday. On Tuesday, she assigned 13 more problems. Over the two days, a tot
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Answer:
Critical value is t = 1.9901
Step-by-step explanation:
We are given the results of the sampling :
In-House Credit Card National Credit Card <u>Sample Size</u> : 32 50
<u>Mean Monthly Purchases </u>: $45.67 $39.87
<u>Standard Deviation </u>: $10.90 $12.47
Also, the managers wished to test whether there is a statistical difference in the mean monthly purchases by customers using the two types of credit cards, using a significance level α of 0.05.
<em>Firstly, we will specify our null and alternate hypothesis;</em>
Let = Mean Monthly Purchases of In-House Credit Card
= Mean Monthly purchases of National Credit Card
So, Null Hypothesis, :
= 0 {means that there is no difference in the mean monthly purchases by customers using the two types of credit cards}
Alternate Hypothesis, :
0 {means that there is statistical difference in the mean monthly purchases by customers using the two types of credit cards}
The test statistics that will be used here is <u>Two-sample t-test statistics</u>;
T.S. = ~
where, = Sample mean Purchases of In-House Credit Card = $45.67
= Sample mean Purchases of National Credit Card = $39.87
= pooled variance
= In-house credit card sample = 32
= National credit card sample = 50
So, degree of freedom of t-value here is (32 + 50 - 2) = 80
Now, at 0.05 significance level, t table gives critical value of t = 1.9901 at 80 degree of freedom.
<em>Therefore, the critical value assuming the population standard deviations are not known but that the populations are normally distributed with equal variances is t = 1.9901.</em>