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tamaranim1 [39]
3 years ago
14

A random sample of 12 supermarkets from Region 1 had mean sales of 84 with a standard deviation of 6.6. A random sample of 17 su

permarkets from Region 2 had a mean sales of 78.3 with a standard deviation of 8.5. Does the test marketing reveal a difference in potential mean sales per market in Region 2? Let μ1 be the mean sales per market in Region 1 and μ2 be the mean sales per market in Region 2. Use a significance level of α=0.02 for the test. Assume that the population variances are not equal and that the two populations are norm
Mathematics
1 answer:
Sladkaya [172]3 years ago
6 0

Answer:

We conclude that there is no difference in potential mean sales per market in Region 1 and 2.

Step-by-step explanation:

We are given that a random sample of 12 supermarkets from Region 1 had mean sales of 84 with a standard deviation of 6.6.

A random sample of 17 supermarkets from Region 2 had a mean sales of 78.3 with a standard deviation of 8.5.

Let \mu_1 = mean sales per market in Region 1.

\mu_2  = mean sales per market in Region 2.

So, Null Hypothesis, H_0 : \mu_1-\mu_2 = 0      {means that there is no difference in potential mean sales per market in Region 1 and 2}

Alternate Hypothesis, H_A : > \mu_1-\mu_2\neq 0      {means that there is a difference in potential mean sales per market in Region 1 and 2}

The test statistics that will be used here is <u>Two-sample t-test statistics</u> because we don't know about population standard deviations;

                            T.S.  =  \frac{(\bar X_1 -\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+ {\frac{1}{n_2}}} }   ~  t__n_1_+_n_2_-_2

where, \bar X_1 = sample mean sales in Region 1 = 84

\bar X_2 = sample mean sales in Region 2 = 78.3

s_1  = sample standard deviation of sales in Region 1 = 6.6

s_2  = sample standard deviation of sales in Region 2 = 8.5

n_1 = sample of supermarkets from Region 1 = 12

n_2 = sample of supermarkets from Region 2 = 17

Also, s_p=\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times  s_2^{2}  }{n_1+n_2-2} }  = s_p=\sqrt{\frac{(12-1)\times 6.6^{2}+(17-1)\times  8.5^{2}  }{12+17-2} } = 7.782

So, <u><em>the test statistics</em></u> =  \frac{(84-78.3)-(0)}{7.782 \times \sqrt{\frac{1}{12}+ {\frac{1}{17}}} }  ~   t_2_7

                                   =  1.943  

The value of t-test statistics is 1.943.

 

Now, at a 0.02 level of significance, the t table  gives a critical value of -2.472 and 2.473 at 27 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so we have<u><em> insufficient evidence to reject our null hypothesis</em></u> as it will not fall in the rejection region.

Therefore, we conclude that there is no difference in potential mean sales per market in Region 1 and 2.

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