At a certain large university a sociologist speculates that male students spend considerably more money on junk food than do female students. To test her hypothesis, the sociologist randomly selects from the registrars' records the names of 200 students. Of these 125 are men and 75are women. The average amount spent on junk food per week by the men is Rs.400/- and standard deviation is Rs100/-. For the women, the sample mean is Rs.450/- and the S.D is Rs.150/-.Test the

Question

At a certain large university a sociologist speculates that male students spend considerably more money on junk food than do female students. To test her hypothesis, the sociologist randomly selects from the registrars' records the names of 200 students. Of these 125 are men and 75are women. The average amount spent on junk food per week by the men is Rs.400/- and standard deviation is Rs100/-. For the women, the sample mean is Rs.450/- and the S.D is Rs.150/-.Test the

difference between the mean at 0.05 level of significance

Answer 1

To test the difference between the mean amount spent on junk food per week by male and female students at a 0.05 level of significance, we can use a two-sample t-test.

Null Hypothesis (H0): There is no significant difference in the mean amount spent on junk food per week by male and female students.
Alternative Hypothesis (H1): Male students spend significantly more money on junk food than female students.

We can calculate the t-test statistic using the formula:

t = (mean1 - mean2) / sqrt((s1^2/n1) + (s2^2/n2))

Where:
- mean1 = 400 (mean amount spent per week by male students)
- mean2 = 450 (mean amount spent per week by female students)
- s1 = 100 (standard deviation for male students)
- s2 = 150 (standard deviation for female students)
- n1 = 125 (number of male students)
- n2 = 75 (number of female students)

Plugging in the values, we get:
t = (400 - 450) / sqrt((100^2/125) + (150^2/75))
t = -50 / sqrt(80 + 300)
t = -50 / sqrt(380)
t = -50 / 19.49
t = -2.57

The degrees of freedom for a two-sample t-test is equal to (n1 + n2 - 2) = (125 + 75 - 2) = 198.

Looking up the critical t-value for a two-tailed test with 198 degrees of freedom and a significance level of 0.05, we get a critical t-value of approximately ±1.973.

Since our calculated t-value (-2.57) is less than the critical t-value (-1.973), we reject the null hypothesis. This means that there is a significant difference in the amount spent on junk food per week between male and female students, with male students spending significantly less on average.