Q1
Null hypothesis (H0H_0H0): The average computer use time for high school juniors is 3.2 hours per day (μ=3.2\mu = 3.2μ=3.2 hours).
Alternative hypothesis (HaH_aHa): The average computer use time for high school juniors is not 3.2 hours per day (μ≠3.2\mu \neq 3.2μ=3.2 hours).
Q2
A. Hypotheses:
- H0H_0H0: The mean battery life is 300 hours (μ=300\mu = 300μ=300).
- HaH_aHa: The mean battery life is not 300 hours (μ≠300\mu \neq 300μ=300).
B. Statistical test: A one-sample t-test is appropriate given the sample size (n=20) and the known sample standard deviation.
C. Significance level and conclusion: Using a significance level of 0.05, calculate the test statistic:
t=xˉ−μs/n=270−30050/20=−2.68t = \frac{\bar{x} - \mu}{s / \sqrt{n}} = \frac{270 - 300}{50 / \sqrt{20}} = -2.68t=s/nxˉ−μ=50/20270−300=−2.68If the calculated t-value exceeds the critical value, we reject H0H_0H0; otherwise, we fail to reject it. Based on this result, we conclude there is enough evidence to reject the CEO's claim at the 0.05 significance level.
Q3
If the difference between the hypothesized population mean and the sample mean is large, we reject the null hypothesis. If the difference is small, we fail to reject the null hypothesis.
Q4
Null hypothesis (H0H_0H0): The mean weight of the parts is 19 pounds (μ=19\mu = 19μ=19).
Alternative hypothesis (HaH_aHa): The mean weight of the parts is not 19 pounds (μ≠19\mu \neq 19μ=19).
Q5
Null hypothesis (H0H_0H0): The average SAT score of the students is 1,020 (μ=1020\mu = 1020μ=1020).
Alternative hypothesis (HaH_aHa): The average SAT score of the students is not 1,020 (μ≠1020\mu \neq 1020μ=1020).