Adults aged 18 years old and older were randomly selected for a survey on obesity. Adults are considered obese if their body mass index (BMI) is at least $30$. The researchers wanted to determine if the proportion of women who are obese in the south is less than the proportion of southern men who are obese. The results are shown in Table 10.27. Test at the 1% level of significance.

	Number who are obese	Sample size
Men	42769	155,525
Women	67169	248,775

Given that, the significance level tested at $1\%$

(a) The null hypothesis is stated as follows: $p_1\le p_2$

(b) The alternate hypothesis is stated as follows:$p_1>p_2$

(c) The disparity between male and female proportions is the random variable.

(d) Two proportions have a normal distribution.

(e) Fill in all requirements using Minitab's two-sample t-test option.

Test and CI for two proportions

Difference is $p\left(1\right)-p\left(2\right)$
Estimate for difference: $0.00499859$
$95$ CI difference: $(0.00217582,0.00782136)$
Difference is$p\left(1\right)-p\left(2\right)$
Estimate for difference:$0.00499859$
95 s CI for difference is (0.00217582,0.00782136)
Test for difference =$0$( vs not =$0$):$2=3.47$

$P$-Value is $0.001$
Fisher's exact test: P-Value$=0.001$
Test statistics$3.47$

(f) $p$-value is $0.001$

(g)

i) $\alpha =0.01$

ii) Decision: Null hypothesis not rejected .

iii) $p>\alpha$.

iv) There is insufficient data to infer that the proportion of males who enjoy shopping for electronic equipment is greater than the proportion of women at the$5\%$ level of significance.