Some studies have shown that in the United States, men spend more than women buy

Question

Some studies have shown that in the United States, men spend more than women buying gifts and cards on Valentine’s Day. Suppose a researcher wants to test this hypothesis by randomly sampling nine men (sample1) and 10 women (sample 2) with comparable demographic characteristics from various large cities across the United States. Each study participant is asked to keep a log beginning one month before Valentine’s Day and record all purchases made for Valentine’s Day during that one-month period. . Use a 1% level of significance to test to determine if, on average, men actually do spend significantly more than women on Valentine’s Day. Assume that such spending is normally distributed in the population and that the population variances are equal.   Data are in the attached file below.

1) What is the calculated value of the test statistic?

2)Using Excel, determine the exact p-value.

3)What is your decision about the null hypothesis?

a. Fail to reject.

b.Reject.

c. Conclude the null is supported.

d.Find the null is true.

a. Fail to reject.

b.Reject.

c. Conclude the null is supported.

d.Find the null is true.

1 Men Women 2 $107.48 $125.98 3 $143.61$ 45.53 4 $ 90.19 $ 56.35 5 $125.53 $ 80.62 6 70.79 $ 46.37 7 83.00 44.34 8 $129.63 $ 75.21 9 $154.22 68.48 10 $ 93.80 $ 85.84 $126.11 3 2

Explanation / Answer

Solution:

Here, we have to use pooled variance t test for the difference between two population means. The null and alternative hypotheses are given as below:

Null hypothesis: H0: There is no any significant difference in the average spends by men and women on Valentine’s Day.

Alternative hypothesis: On an average, men do spend significantly more than women on Valentine’s Day.

H0: µ1 = µ2 vs. Ha: µ1 > µ2 (one tailed – upper/right tailed test)

We are given a level of significance = ? = 0.01

Test statistic formula is given as below:

t = (X1bar – X2bar) / sqrt[Sp2*((1/n1)+(1/n2))]

Where Sp2 is pooled variance

Sp2 = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

From given data, we have

X1bar = 110.9167

S1 = 28.79159

n1 = 9

X2bar = 75.483

S2 = 30.50778

n2 = 10

df = n1 + n2 – 2 = 9 + 10 – 2 = 17

Sp2 = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

Sp2 = [(9 – 1)*28.79159^2 + (10 – 1)*30.50778^2]/(9 + 10 – 2)

Sp2 = 882.8333

t = (X1bar – X2bar) / sqrt[Sp2*((1/n1)+(1/n2))]

t = (110.9167 – 75.483) / sqrt[882.8333*((1/9)+(1/10))]

t = 35.4337 / 13.6520

t = 2.5955

P-value = 0.0094 (by using t-table/excel)

? = 0.01

P-value < ? = 0.01

So, we reject the null hypothesis that There is no any significant difference in the average spends by men and women on Valentine’s Day.

There is sufficient evidence to conclude that the men spend more than women on Valentine’s Day.

Answers:

Question 1:

Calculated value of test statistic = t = 2.5955

Question 2:

P-value = 0.0094

Question 3:

b. Reject

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