2. 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?
3. 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.
4. 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.
5. 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%
7. Minitab output
Test and CI for Two Proportions -
confidence interval
Sample X N Sample p
1 165 189 0.873016
2 257 298 0.862416
Difference = p (1) - p (2)
Estimate for difference: 0.0105998
95% CI for difference: (-0.0509046,
0.0721041)
Test for difference = 0 (vs not = 0):
Z = 0.34
P-Value = 0.736
Fisher's exact test: P-Value = 0.786
Test and CI for Two Proportions –
hypothesis
Sample X N Sample p
1 165 189 0.873016
2 257 298 0.862416
Difference = p (1) - p (2)
Estimate for difference: 0.0105998
95% upper bound for difference:
0.0622158
Test for difference = 0 (vs < 0):
Z = 0.34
P-Value = 0.632
Fisher's exact test: P-Value = 0.679
8. Checking the normal
model
Males
NP ≥ 10
189(0.873) = 164.997
164.997 ≥ 10 ✓
N(1-P) ≥ 10
189(1-0.873) = 24.003
24.003 ≥ 10 ✓
Females
NP ≥ 10
298(0.862) = 256.876
256.876 ≥ 10 ✓
N(1-P) ≥ 10
298(1-0.862) = 41.124
41.124 ≥ 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
9. 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.
10. 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
11. 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.
12. Null hypothesis
Defining 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 = 0.0105998
Null 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 = 0
Ha: 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
13. 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.
