A sample of 25 different payroll departments found that the employees worked an average of 310.3 days a year with a standard deviation of 23.8 days. What is the 95% confidence interval for the average days worked of all payroll departments? Assume that the distribution of days worked is approximately normally distributed.
Answer (300.47536 , 320.12464)
The Pizza Shop wanted to determine what proportion of its customers ordered only cheese pizza. Out of 80 customers surveyed, 15 ordered cheese pizza. What is the 95% confidence interval of the true proportion of customers who order only cheese pizza?
Answer: (0.102,0.273)
A real estate agent wishes to estimate the average size of homes for sale in his/her state. From a study nationwide, the mean was found to be 2515 ft2 and the standard deviation was found to be 936.87 ft2. In order to be 95% confident and accurate within 50 square feet of the true mean for this real estate agent's U.S. state, what is the minimum sample size required?
Answer: 937
An economics professor randomly selected 8 millionaires in the United States. The average age of these millionaires was 54.8 years with a standard deviation of 7.9 years. Assume that the ages of millionaires follow a normal distribution. What is the 95% confidence interval for the mean age of all United States millionaires?
Answer: (48.1944,61.4056)
A recent gasoline survey said that the national average price of gasoline was $1.498 a gallon. It was felt that gasoline in Texas was significantly lower than the national average. A survey of 10 different suburbs in Dallas, Texas found the average price of gasoline to be $1.294 a gallon with a standard deviation of $0.0526. Using a 0.05 significance level, test this suspicion by answering parts a-d below.
a)State the null and the alternative hypotheses by filling in the blanks.
H0: ≥ 1.498 vs. H1: < 1.498
b) State the distribution you use to find the critical value(s) and state the critical value(s).
t-distribution, 9 df
Critical value = -1.833
c) Calculate the test statistic value. -12.264
d) State your decision to reject or fail to reject Ho, and state your conclusion regarding the researcher's claim.
Reject Ho, using a 0.05 significance we can conclude that gasoline in Texas was significantly lower than the national average
A 2005 report from the Medical Association stated that the average salary of psychiatrists was $189,121 with a standard deviation of $26,975. A random sample of 64 psychiatrists was recently taken to test the claim that the average salary of psychiatrists is still $189,121, and it was found that the p-value is 0.036. Answer the following:
a. State a significance level for which the appropriate conclusion would be that the current average salary of all psychiatrists is not significantly different from $189,121.
Answer: We can take any significance level less than the p-value (0.036), for example
we can take = 0.02
b. If using the traditional hypothesis testing procedure resulted in a decision to accept Ho and the null hypothesis turned out to actually be false, then which of the following errors could have been made: margin of error, sampling error, standard error, Type I error, Type II error?
Answer: Type II error
It has been claimed that at U.C.L.A. fewer than 40% of the students live on campus. From a sample of 250 students, 90 live on campus. Using a significance level of 0.05, test this claim by answering parts a-d below.
a) State the null and the alternative hypotheses by filling in the blanks.
H0: p ≥ 0.4 vs. H1:p < 0.4
b) State the distribution you use to find the critical value(s) and state the critical value(s).
Answer: z-distribution (or standard normal)
c) Calculate the test statistic value.
Answer: -1.291
d) State your decision to reject or fail to reject Ho, and state your conclusion regarding the researcher's claim.
Answer:
Critical value = -1.645
Rejection region: {z/z< -1.645}
Since the statistical value (-1.291) is greater that the critical value (-1.645) we do not reject Ho
Using = 0.05 we do not have enough evidence to say that fewer than 40% of the students live on campus
A doctor believes that the variance of systolic blood pressure is 450. A random sample of 24 patients found an average systolic blood pressure of 119 with a standard deviation of 22.8. Assume the variable is normally distributed and a = 0.05. Answer parts a-e below.
a) State the null and the alternative hypotheses by filling in the blanks.
H0: 2 = 450 vs. H1: 2450
b) State the distribution you use to find the critical value(s) and state the critical value(s).
Distribution: Chi-square, df = 23
c) Calculate the test statistic value.
Answer: 26.570
d) If we were to test a claim that the variance in systolic blood pressure is higher than the variance in diastolic blood pressure, which distribution would we use to test this claim?
Answer: Chi-square(df = 23)
Is missing part e?
Imported Wine Domestic Wine
Mauricio Cruz, a wine merchant for Cruz's Spirits Emporium, wants to determine if there is an appreciable difference between the average price of imported wine and the average price of domestic wine. The data obtained is shown below. Using a 5% significance level, conduct this test of hypotheses by answering parts a-d below.
Imported Wine Domestic Wine
Xbar = 8.03 Xbar = 7.59
s = 2.51 s = 3.26
n = 41 n = 31
a) State the null and the alternative hypotheses by filling in the blanks.
H0:1 - 2 = 0 vs. H1: 1- 2 0
b) State the distribution you use to find the critical value(s) and state the critical value(s).
t-distribution: df = 70
Critical values: -1.994 and 1.994
c) Calculate the test statistic value.
Statistic: 0.638
d) State your decision to reject or fail to reject Ho, and state your conclusion regarding the claim.
Rejection region: {t/ t<-1.994 or t > 1.994}
Since the stastistic value (0.638) is not in the rejection region (fail to reject Ho)
Using = 0.05 we do not have enough evidence to say that there is an appreciable difference between the average price of imported wine and the average price of domestic wine
- A researcher claims that the average miles per gallon (mpg) using an octane booster is higher than the average mpg without an octane booster. A random sample of five cars was selected; the cars were driven for two weeks without the booster and two weeks with the booster. Data are provided below. Test this claim, using a = 0.05, by answering parts a-d.
mpg Without Booster mpg With Booster
21.2 23.8 25.4 25.4 20.9 22.4
27.6 28.3
21.8 21.5
a) State the null and the alternative hypotheses by filling in the blanks.
H0: 1-2≥ 0vs. H1: 1-2 < 0
b) State the distribution you use to find the critical value(s) and state the critical value(s).
t-distribution, df = 4
Critical value = -2.776
c) Calculate the test statistic value.
Statistic = t = -1.710
d) State your decision to reject or fail to reject Ho, and state your conclusion regarding the researcher's claim.
Rejection region: {t/ t<-2.776 }
Since the stastistic value (-1.710) is not in the rejection region (we fail to reject Ho)
Using = 0.05 we do not have enough evidence to say that that the average miles per gallon (mpg) using an octane booster is higher than the average mpg without an octane booster