Which of the following is a correct cosine ratio for the figure?

Here a "success" is "obtaining a 7 or a 5." Assuming that the 10-sided die is marked {1, 2, 3, ... , 9, 10}, there are 2 ways in which to get a "success" out of 10 possible outcomes. Thus, P(5 or 7) is 2/10, or 1/5, or 02.

5 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 Sample Size : 32 50

Mean Monthly Purchases : $45.67 $39.87

Standard Deviation : $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.

Firstly, we will specify our null and alternate hypothesis;

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 Two-sample t-test statistics;

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.

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.

4 0

3 years ago