Question

please help me fix this issue, using the =t.inv on excel is giving me -2.406

can -2.047 be a typo??

thanks 

Problem 3. According to a recent study, Americans get an average of 7 hours of sleep
per night. A random sample of 49 residents in Carrollton, GA has a mean of 6.8 hours
sleep time per night. The sample standard deviation is 0.9 hours. Is it reasonable to
conclude that residents in Carrollton have less sleep than the typical American?
a. Setup the appropriate null and alternative hypotheses.
b. Construct the relevant test statistic.
c. Test the hypotheses at the 0.01 significance level and state your conclusion.
Let u be the mean sleep time among Carrollton residents. The two hypotheses are
Ho:μ ≥7 and HA: <7. The test statistic is given by
t=
另一位
s/√
6.8-7
0.9/√49
-1.556.
This is a left-sided t test with a degrees of freedom of n-1=49-1 = 48. The one-
sided critical value is -2.047 (i.e., we reject if t < -2.047). We thus do not reject
the null hypothesis and cannot establish the conclusion that Carrollton residents
sleep less than the typical American.
expand button
Transcribed Image Text:Problem 3. According to a recent study, Americans get an average of 7 hours of sleep per night. A random sample of 49 residents in Carrollton, GA has a mean of 6.8 hours sleep time per night. The sample standard deviation is 0.9 hours. Is it reasonable to conclude that residents in Carrollton have less sleep than the typical American? a. Setup the appropriate null and alternative hypotheses. b. Construct the relevant test statistic. c. Test the hypotheses at the 0.01 significance level and state your conclusion. Let u be the mean sleep time among Carrollton residents. The two hypotheses are Ho:μ ≥7 and HA: <7. The test statistic is given by t= 另一位 s/√ 6.8-7 0.9/√49 -1.556. This is a left-sided t test with a degrees of freedom of n-1=49-1 = 48. The one- sided critical value is -2.047 (i.e., we reject if t < -2.047). We thus do not reject the null hypothesis and cannot establish the conclusion that Carrollton residents sleep less than the typical American.
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