Gender vs. Smartphone use

Kris Rogers

To indicate if the proportion of male smart phone users is significantly lower than the proportion of female smart phone users.

What's the purpose of our study?

Demographic We obtained our data set from the spring 2013Math 075

Class Survey. Our data set consists of 289 Females and 189 Males. The original total number of data was 505 students; we

threw out 18 pieces of data due to incomplete responses on the original survey questions, “What is your gender?” “Do you use a smartphone?”

After eliminating the failed responses, our data set now contains 487 075 Math students from spring 2013.

Population

The population can’t be the general population of COC because math 075 isn't necessarily taken by every student attending College of the Canyons. Students may have placed in a higher math class. However, we can make inferences on liberal arts majors at COC, only because we are unaware if other colleges math 075 is equivalent to COC’s.

Survey Results Survey question: Do you own a smartphone yes or no? Students who use a smartphone

Females—257/298 p-hat = 0.862 or 86.2% Males—165/189 p-hat = 0.873 or 87.3%

Students who do not use a smartphone Females—41/298 p-hat = 0.138 or 13.8% Males—24/189 p-hat = 0.127 or 12.7%

Survey Results

Minitab outputTest and CI for Two Proportions - confidence intervalSample X N Sample p1 165 189 0.8730162 257 298 0.862416Difference = p (1) - p (2)Estimate for difference: 0.010599895% CI for difference: (-0.0509046, 0.0721041)Test for difference = 0 (vs not = 0): Z = 0.34 P-Value = 0.736Fisher's exact test: P-Value = 0.786

Test and CI for Two Proportions – hypothesis Sample X NSample p1 165 189 0.8730162 257 298 0.862416Difference = p (1) - p (2)Estimate for difference: 0.010599895% upper bound for difference: 0.0622158Test for difference = 0 (vs < 0): Z = 0.34 P-Value = 0.632Fisher's exact test: P-Value = 0.679

Checking the normal model

MalesNP ≥ 10189(0.873) = 164.997164.997 ≥ 10 ✓N(1-P) ≥ 10189(1-0.873) = 24.00324.003 ≥ 10 ✓

FemalesNP ≥ 10298(0.862) = 256.876256.876 ≥ 10 ✓N(1-P) ≥ 10298(1-0.862) = 41.12441.124 ≥ 10 ✓

The sample is assumed to be random, and for both males and females the values for successes and failures is greater than ten. Thus, the conditions have been satisfied and we can proceed using the normal model.

Interpretation of confidence interval

• We used Minitab to receive the 95% confidence interval of (-0.0509046, 0.0721041). When interpreting the interval we changed the decimals to percentages giving us 5.1% and 7.2%.

We are 95% confident that the proportion of males who use a smartphone is between 5.1% less and 7.2% more that the proportion of females who use a smartphone.

Test-statistic interpretation

• Minitab gives the Z-Score as Z = 0.34

The observed difference in sample proportions of 0.0105998 is about 0.34 standard errors above the assumed difference of zero

P-value interpretation• Minitab gives the p-value as 0.632

Assuming the null hypothesis is true, the probability that random samples will have a difference on sample proportions that is at least as extreme as 0.0105998 is 0.

Null hypothesisDefining P1 and P2• P1 = Pmale = Pm: The proportion of males who use a smartphone.• P2 = Pfemale = Pf: The proportion of females who use a

smartphone.The mean of the distribution of sample differences is P1-P2 or Pm-Pf• Estimate for difference: 0.873016-0.862416 = 0.0105998Null Hypothesis Ho: The proportion of male Liberal Arts major who use a smart phone is equal to the proportion of female Liberal Arts majors who use a smart phone. Pm-Pf = 0Ha: The proportion of male Liberal Arts major who use a smart phone is less than the proportion of female Liberal Arts majors who use a smart phone. Pm-Pf < 0

Conclusion

Since our P-value of 0.632 is greater than our significance level of 0.05, we fail to reject the null hypothesis and do not accept the alternative hypothesis. The data from thus study does not provide enough evidence to support that male smart phone users is less than female smartphone users.