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EastWind [94]
2 years ago
9

42 points!

Mathematics
2 answers:
kupik [55]2 years ago
5 0

Answer:

C

Step-by-step explanation:

30x20=600

Nimfa-mama [501]2 years ago
4 0
C) $600 because 30*20 is 600
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3 0
2 years ago
CAN SOMEONE PLS HELP!!!
Soloha48 [4]
Where’s the question
8 0
3 years ago
A major retail clothing store is interested in estimating the difference in mean monthly purchases by customers who use the stor
mixer [17]

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 \mu_1 = Mean Monthly Purchases of In-House Credit Card

     \mu_2 = Mean Monthly purchases of National Credit Card

So, Null Hypothesis, H_0 : \mu_1-\mu_2 = 0  {means that there is no difference in the mean monthly purchases by customers using the two types of credit cards}

Alternate Hypothesis, H_0 : \mu_1-\mu_2\neq 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. = \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \sqrt{\frac{1}{n_1}+\frac{1}{n_2}  } }  ~ t_n___1+n_2-_2

where, \bar X_1 = Sample mean Purchases of In-House Credit Card = $45.67

           \bar X_2 = Sample mean Purchases of National Credit Card = $39.87

            s_p = pooled variance

            n_1 = In-house credit card sample = 32

            n_2 = 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>

4 0
2 years ago
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