Xr010-29
Descriptive Statistics
Times
Valid
15
Missing
0
Mean
147.333
95% CI Mean Upper
175.040
95% CI Mean Lower
119.627
Std. Deviation
50.031
Minimum
60.000
Maximum
240.000
The 95% confidence interval estimate for the population mean, based on the sample data provided, is approximately
from 119.63 to 175.04.
10-51
In this case, the professor wants a 95% confidence interval and a margin of error of 2 marks (E). The population
standard deviation is assumed to be 12 (σ).
The z-score for a 95% confidence level is commonly 1.96 (z) for a two-tailed test from the standard normal
distribution.
n = ((1.96 x 12)/2)^2 = 138.2976
To estimate the mean mark from 25 years ago to within 2 marks with 95% confidence, assuming the population
standard deviation is 12, the professor should take a sample of approximately 138 students' marks.
Xr11-45
One Sample T-Test
One Sample T-Test
t
PerChange
df
-1.676
p
49
0.100
Note. For the Student t-test, the alternative hypothesis specifies that the mean is different from 0.
Note. Student's t-test.
The one-sample t-test yields a t-statistic of approximately -1.68 with a p-value of approximately 0.100. Since the pvalue is greater than the significance level of 0.10, we do not have sufficient evidence to reject the null hypothesis at
the 10% significance level. This means we cannot conclusively infer that the new safety equipment is effective based
on this data.
Xr12-21
Descriptive Statistics
Winnings
Valid
Missing
Mean
95% CI Mean Upper
95% CI Mean Lower
Std. Deviation
Minimum
Maximum
15
0
24050.667
33678.784
14422.549
17386.129
990.000
52820.000
The mean winnings of all "Jeopardy!" game players, based on the provided sample, is estimated at
approximately 24051. The 95% confidence interval for the mean winnings ranges from about 14423 to 33679.
Xr12-31
Descriptive Statistics
Commercials
Valid
116
Missing
0
Mean
15.267
Std. Deviation
5.723
Minimum
8.000
Maximum
37.000
One Sample T-Test
One Sample T-Test
t
Commercials
df
0.503
p
115
0.308
Note. For the Student t-test, the alternative hypothesis specifies that the mean is greater than 15.
Note. Student's t-test.
Assumption Checks
Test of Normality (Shapiro-Wilk)
W
Commercials
0.877
p
< .001
Note. Significant results suggest a deviation from normality.
The analysis of the data with a one-sample t-test reveals that there isn't sufficient evidence to conclude
that the mean number of special Super Bowl ads created by firms is greater than 15. The sample mean is
approximately 15.27, but the statistical test resulted in a p-value of approximately 0.308, which is too high
to indicate a significant difference from 15.
Question: Can I use Students’ t-test if the data is not normally distributed?