9.
right-tailed test; μ
10.
left-tailed test; p
11.
two-tailed test; σ
12.
right-tailed test; p
13.
left-tailed test; μ
14.
two-tailed test; σ
15.
Ho: p = 0.105
H1: p > 0.105
type I error: The researcher concludes that the proportion of registered births in the US to teenage mothers has increased since 2007, when in truth the proportion had not increased.
type II error: The researcher concludes that the proportion of registered births in the US to teenage mothers has not increased since 2007, when in truth the proportion had increased.
17.
Ho: μ = $218,600
H1: μ < $218,600
type I error: A real estate broker claims that the mean price of an existing single-family home in 2009 has decreased since then, when in truth the mean price has not decreased.
type II error: A real estate broker claims that the mean price of an existing single-family home in 2009 has not decreased since then, when in truth the mean price has decreased.
19.
Ho: σ = 0.7 psi
H1: σ < 0.7 psi
type I error: The quality-control manager rejects the hypothesis that 0.7 psi is the pressure variability required to open the valve, when in truth the pressure variability is 0.7 psi.
type II error: The quality-control manager does not reject the hypothesis that the pressure variability required is 0.7 psi, when in truth the pressure variability is less than 0.7 psi.
21.
Ho: μ = $47.47
H1: μ ≠ $47.47
type I error: A researcher concludes that the mean monthly phone bill today is different from 2007, when in fact the mean monthly phone bill has not changed.
type II error: A researcher concludes that the mean monthly phone bill today is the same as in 2007, when in fact the mean monthly phone bill has changed.
7.
9.
11.
13.
A P-value of 0.2743 means that there is a 0.2743 probability that a sample will result in a sample proportion as high or higher than the one obtained if the population proportion is actually 0.5. We do not reject the null hypothesis because the probability is not low. There isn’t enough evidence to conclude that the dart-picking strategy resulted in a majority of the winners.
15.
P-value = 0.5832
There is little evidence at the α = 0.01 level of significance to reject the null hypothesis and conclude that more than 1.9% of Lipitor users experience flulike symptoms as a side effect.
17.
P-value = 0.1492
There is little evidence at the α = 0.05 level of significance to reject the null hypothesis and conclude that a majority of adults in the United States believe they will not have enough money in retirement.
19.
P-value = 0.0047
There is strong evidence at the α = 0.05 level of significance to support the hypothesis that the proportion of employed adults who feel basic mathematical skills are critical or very important to their job is greater than 0.56.
1.
2.602
-1.72989
±2.179
2.
1.321
-2.426
±2.708
3.
-1.379
-1.714
Skip this problem.
There is little evidence for the researcher to reject the null hypothesis because the test statistic is greater than the critical value.
4.
2.644
2.492
Skip this problem.
There is enough evidence for the researcher to reject the null hypothesis because the test statistic is greater than the critical value.
5.
2.502
±2.819
Skip this problem.
There isn’t enough evidence for the researcher to reject the null hypothesis because the test statistic is between the critical values.
LB: 99.39, UB: 110.21. We do not reject the null hypothesis because 100 is included in the 99% confidence interval.
11.
Ho: μ = 67
Ha: μ > 67
A P-value of 0.02 means that there is a 0.02 probability of calculating a sample mean of $73 or more from a population with a mean of $67.
The P-value is lower than the level of significance so we reject the null hypothesis. There is enough evidence to conclude that the mean withdrawal amount from a PayEase ATM is more than the mean withdrawal amount from a standard ATM.
13.
Ho: μ = 22
Ha: μ > 22
The sample size is large and the sample is random. It’s assumed that the scores are independent because the sample is smaller than the population.
The test statistic is greater than the critical value so we reject the null hypothesis. t0 = 2.176 > t0.05 = 1.660
There is enough evidence to conclude that the students are scoring above 22 on the math portion of the ACT.
15.
t0 = -4.553
There is enough evidence to reject the null hypothesis and conclude that the mean hippocampal volume in alcoholic adolescents is less than the normal mean volume of 9.02 cm3.
17.
t0 = 0.813
There isn’t enough evidence at the α = 0.05 level of significance to conclude that high-income individuals have higher FICO scores.
19.
LB: 35.44, UB: 42.36
There isn’t enough evidence to conclude that the mean age of a death-row inmate has changed since 2002 because the interval includes 40.7 years.
1.
28.869
14.041
16.047, 45.722
3.
20.496
13.091
Skip this problem
We do not reject the null hypothesis because the test statistic is greater than the critical value.
x <- sd(c(108.5,80.4,67.4,95.5,58.0,86.3,75.9,70.9,65.1,72.0))
x## [1] 15.2050413.
Note. The sample standard deviation of the data shown is.\(15.205043\). You will have to knit this to see the number though.
6.422
There isn’t enough evidence at the α = 0.05 level of significance to conclude that the standard deviation wait-time is less than 18.0 seconds.
15.
15.639
There is enough evidence at the α = 0.10 level of significance to conclude that Rose is a more consistent player than other shooting guards in the NBA.