A random sample of 100 people from region A and a random sample of 100 people from region B were surveyed about their grocery-shopping habits. From the region A sample, 16 percent of the people indicated that they shop for groceries online. From the region B sample, 24 percent of the people indicated that they shop for groceries online. At the significance level of α=0.05, do the data provide convincing statistical evidence that there is a difference between the two regions for the population proportions of people who shop online for groceries? Complete the appropriate inference procedure to support your answer.

Step-by-step explanation:

This is a test of 2 population proportions. Let a and b be the subscript for people from region A and region B. The population proportions would be pa and pb

pa - pb = difference in the proportion of people from region A and region B.

The null hypothesis is

H0 : pa = pb

pa - pb = 0

The alternative hypothesis is

H1 : pa ≠ pb

pa - pb ≠ 0

it is a two-tailed test

Sample proportion = x/n

Where

x represents number of success

n represents number of samples

For region A,

xa = 16

na = 100

pa = 16/100 = 0.16

For region B,

xb = 24

nb = 100

pb = 24/100 = 0.24

The pooled proportion, pc is

pc = (xa + xb)/(na + nb)

pc = (16 + 24)/(100 + 100) = 0.2

1 - pc = 1 - 0.2 = 0.8

z = (pa - pb)/√pc(1 - pc)(1/na + 1/nb)

z = (0.16 - 0.24)/√(0.2)(0.8)(1/100 + 1/100) = - 0.08/√0.0002891223

z = - 1.41

Since it is a 2 tailed test, we would find the p value by doubling the area to the left of the z score to include the area to right.

Area to the left from the normal distribution table is

0.07927

P value = 0.07927 × 2 = 0.16

Since 0.05 < 0.16, we would accept the null hypothesis

Therefore, the data does not provide convincing statistical evidence that there is a difference between the two regions for the population proportions of people who shop online for groceries.

55/i dont know i do it for the points

step-by-step explanation:

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