Author Question: A study has been conducted to determine whether the mean spending for recreational activities during ... (Read 116 times)

melly21297 · **on:** Jun 24, 2018

A study has been conducted to determine whether the mean spending for recreational activities during the month of August differs for residents of three cities.

Random samples of 30 people were selected from each city and their spending on recreation was recorded during August. The following output was generated using Excel:ANOVA: Single FactorSUMMARYGroups Count Sum Average VarianceCity 1 30 7897.179 236.2393 3334.11City 2 30 10322.1 344.0701 2201.818City 3 30 6045.102 201.5034 2215.919ANOVASource of Variation SS df MS F P-value F critBetween Groups 306701.8 2 153350.9 59.3475 5.54E-17 3.101292Within Groups 224803.5 87 2583.949 Total 531505.4 89 Based on the information provided, should we conclude that the three populations (cities) have equal mean spending during August? Test at the 0.05 level of significance.

Question 2

If a binomial distribution applies with a sample size of n = 20, find the probability of 5 successes if the probability of a success is 0.40.

A) 0.1246
B) 0.1286
C) 0.0746
D) 0.0866

millet · **Reply #1 on:** Jun 24, 2018

Answer to Question 1

The null and alternative hypotheses to be tested are:
H0 : 1 = 2 = 3
Ha : not all j are equal.
Since three populations are involved and the samples that have been selected are independent, if we assume the following:
1. The populations are normally distributed.
2. The populations have equal variances.
3. The observations are independent.
Then the one-way analysis of variance test can be used to test the null hypothesis. The test statistic for this test is provided in the Excel output to be F = 59.3475. This value is compared to the F-critical value for 2 and 87 degrees of freedom, which is 3.10.
Since F = 59.3475 > 3.10, we reject the null hypothesis and conclude that not all means are equal. We could also use the p-value approach to conduct the test. Since the p-value is shown to be virtually zero, which is less than alpha = 0.05, we would reject the null hypothesis.

Answer to Question 2

C