Respuesta :
Answer:
(a) Decision rule for 0.01 significance level is that we will reject our null hypothesis if the test statistics does not lie between t = -2.651 and t = 2.651.
(b) The value of t test statistics is 1.890.
(c) We conclude that there is no difference in the mean number of times men and women order take-out dinners in a month.
(d) P-value of the test statistics is 0.0662.
Step-by-step explanation:
We are given that a recent study focused on the number of times men and women who live alone buy take-out dinner in a month.
Also, following information is given below;
Statistic : Men Women
The sample mean : 24.51 22.69
Sample standard deviation : 4.48 3.86
Sample size : 35 40
Let [tex]\mu_1[/tex] = mean number of times men order take-out dinners in a month.
[tex]\mu_2[/tex] = mean number of times women order take-out dinners in a month
(a) So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu_1-\mu_2[/tex] = 0 {means that there is no difference in the mean number of times men and women order take-out dinners in a month}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu_1-\mu_2\neq[/tex] 0 {means that there is difference in the mean number of times men and women order take-out dinners in a month}
The test statistics that would be used here Two-sample t test statistics as we don't know about the population standard deviation;
T.S. = [tex]\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] ~ [tex]t__n_1_-_n_2_-_2[/tex]
where, [tex]\bar X_1[/tex] = sample mean for men = 24.51
[tex]\bar X_2[/tex] = sample mean for women = 22.69
[tex]s_1[/tex] = sample standard deviation for men = 4.48
[tex]s_2[/tex] = sample standard deviation for women = 3.86
[tex]n_1[/tex] = sample of men = 35
[tex]n_2[/tex] = sample of women = 40
Also, [tex]s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }[/tex] = [tex]\sqrt{\frac{(35-1)\times 4.48^{2}+(40-1)\times 3.86^{2} }{35+40-2} }[/tex] = 4.16
So, test statistics = [tex]\frac{(24.51-22.69)-(0)}{4.16 \sqrt{\frac{1}{35}+\frac{1}{40} } }[/tex] ~ [tex]t_7_3[/tex]
= 1.890
(b) The value of t test statistics is 1.890.
(c) Now, at 0.01 significance level the t table gives critical values of -2.651 and 2.651 at 73 degree of freedom for two-tailed test.
Since our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.
Therefore, we conclude that there is no difference in the mean number of times men and women order take-out dinners in a month.
(d) Now, the P-value of the test statistics is given by;
P-value = P( [tex]t_7_3[/tex] > 1.89) = 0.0331
So, P-value for two tailed test is = 2 [tex]\times[/tex] 0.0331 = 0.0662
Using the t-distribution, we have that:
a) Reject H0 if t < -2.645 or t > 2.645.
b) The value of the test statistic is t = 1.871.
c) Since the value of the test statistic is t = 1.871 < 2.645, the decision is made to not reject the null hypothesis.
d) The p-value is of 0.0654.
item a:
At the null hypothesis, it is tested if there is no difference, hence:
[tex]H_0: \mu_M - \mu_W = 0[/tex]
At the alternative hypothesis, it is tested if there is difference, hence:
[tex]H_1: \mu_M - \mu_W \neq 0[/tex]
We need to find the critical value for a two-tailed test, as we are testing if the values are different, with 35 + 40 - 3 = 73 df and a significance level of 0.01.
- Using a calculator, the critical value is [tex]t = \pm 2.645[/tex], hence:
Reject H0 if t < -2.645 or t > 2.645.
Item b:
The standard errors are:
[tex]s_M = \frac{4.48}{\sqrt{35}} = 0.7573[/tex]
[tex]s_W = \frac{3.86}{\sqrt{40}} = 0.6103[/tex]
The distribution of the difference has:
[tex]\overline{x} = \mu_M - \mu_W = 24.51 - 22.69 = 1.82[/tex]
[tex]s = \sqrt{s_M^2 + s_W^2} = \sqrt{0.7573^2 + 0.6103^2} = 0.9726[/tex]
The test statistic is:
[tex]t = \frac{\overline{x} - \mu}{s}[/tex]
In which [tex]\mu = 0[/tex] is the value tested at the null hypothesis, hence:
[tex]t = \frac{\overline{x} - \mu}{s}[/tex]
[tex]t = \frac{1.82 - 0}{0.9726}[/tex]
[tex]t = 1.871[/tex]
The value of the test statistic is t = 1.871.
Item c:
Since the value of the test statistic is t = 1.871 < 2.645, the decision is made to not reject the null hypothesis.
Item d:
Using a t-distribution calculator, with a two-tailed test with t = 1.871 and 73 df, the p-value is of 0.0654.
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