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

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

(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) $\mathrm{Î\pm }=0.01$

ii) Decision: Null hypothesis not rejected .

iii) $p>\mathrm{Î\pm }$.

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.