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Varvara68 [4.7K]
2 years ago
11

What is the slope and the y-intercept.

Mathematics
1 answer:
storchak [24]2 years ago
5 0

Answer:

slope = 1

y-intercept = 5

Step-by-step explanation:

change in y-values = 1

change in x-values = 1

slope = 1/1 or 1

to find the y-intercept ('b'), select an ordered pair:  (1, 6)

use the slope-intercept formula and solve for 'b':

6 = 1(1) + b

6 = 1 + b

5 = b

slope = 1

y-intercept = 5

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A major retail clothing store is interested in estimating the difference in mean monthly purchases by customers who use the stor
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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

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<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>

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