The adoption of smart phones has been discussed a lot. One aspect of these devic

Question

The adoption of smart phones has been discussed a lot. One aspect of these devices is that students tend to be distracted by their phones during class. A recent survey reported that 76% of 1004 randomly sampled females used their mart phones during class time, compared to 71% of 1009 randomly sampled males. Do these results confirm a higher smart phone usage by females?

a) State and test an appropriate hypothesis and state your conclusions. Show your work.

b) Find a 99% confidence interval for the difference in the proportion of males and females who use their phone during class. Interpret your interval.

Please show work to help me understand. Thanks

Explanation / Answer

a) H0: p1 = p2

    H1: p1 > p2

Pooled proportion (P) = (p1 * n1 + p2 * n2)/(n1 + n2)

                                   = (0.76 * 1004 + 0.71 * 1009)/(1004 + 1009)

                                    = 0.7349

SE = sqrt(P * (1 - P) * (1/n1 + 1/n2))

       = sqrt(0.7349 * (1 - 0.7349) * (1/1004 + 1/1009))

       = 0.02

The test statistic z = (p1 - p2)/SE

                              = (0.76 - 0.71)/0.02 = 2.5

P-value = P(Z > 2.5)

             = 1 - P(Z < 2.5)

              = 1 - 0.9938 = 0.0062

At 5% significance level we can conclude that, the P-value is is less than the significance level (0.0062 < 0.05), so the null hypothesis is rejected.

So we can conclude that a higher smarter phonnes usage by females than males.

b) At 99% confidence interval the critical value is z0.99 = 2.33

The 99% confidence interval is

(p1 - p2) + z0.99 * SE

= (0.76 - 0.71) + 2.33 * 0.02

= 0.05 + 0.0466

= 0.0966

As 0 is less than the Upper bound, so the null hypothesis is rejected.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.