Class worksheet #4
- Estimation of the mean (unknown variance), the variance, the proportion
- Hypotheses tests of the mean (unknown variance), the variance, the proportion
1.A study is made by the school Dean’s office regarding the amount of time students spend studying. An argument was investigated that on the average, a student studies more than 40 hour in preparations for the finals. Can you infer from the sample provided below that this is a sound argument at 10% significance level?
- The Baines Investment Company (BIC) is interested in the average number of mutual funds its customers have in their portfolios. A random sample of investors is selected and the number of mutual funds recorded. See file BIC.
(a)Is there sufficient evidence in the sample to support the hypothesis that the number of mutual funds per portfolio is more than 2 on the average? Use 10% significance-level.
(b)In an effort to diversify its customers’ portfolios, BIC has been encouraging its agents to advise its customers to have more than 2 mutual funds in their portfolios. Management is offering a commission to its agents if the average number of mutual funds is proven to be more than 2. With the sample results available, what is the actual probability the commission will be paid erroneously?
(c)What is the probability that the commission will not be paid although the mean number of mutual funds per portfolio is 2.5? Use the sample provided to answer.
- For years, management of High-Desert Bank (HDB) believed the mean loan amount for its commercial-retailcustomers was $60,000. Management expressed concerns about the (too low) amount borrowed by customers. An advertising promotion that was intended to raise the mean loan amounts by more than $500 was conducted. A couple of weeks ago, an intern from the nearby college collected a random sample of customers (commercial and retail). These data are stored in file HDB.
(a)The loan manager wishes to re-estimate the mean loan amount for the commercial customers of his bank. Provide an estimate at 98% confidence level.
(b)Test at 1% significance level the success of the promotion operation in achieving its objective?
(c)The marketing company who conducted the advertising and promotion was promised a bonus if the proportion of customers borrowing more than $65,000 is more than 45%. Should the bonus be paid? (Use = .05).
(d)Management wishes to estimate the proportion of customers who borrow more than $65,000 to within .05 at 90% confidence. What sample size is required?
Solutions
1.(a)The parameter tested is the mean time of study before the final. We need to test the following hypotheses:
H0: = 40
H1: > 40
The rejection region is Running the data we have , so, to find the critical value note that After the standard normal transformation we have, or
Since 42.1 is not greater than 45.66, there is insufficient evidence in the sample to reject the null hypothesis. We cannot conclude at 10% significance level that on the average students study more than 40 hours for the finals.
2.a.We need to set up two hypotheses as follows:
H0: = 2
H1: > 2
Since the standard deviation is unknown we should use the t-distribution to perform the test. The t test is tsample > t.10 where t has 199 degrees of freedom because the sample size n = 200.
We can use Excel (the TINV function) to find that t.10 = 1.2858; (You can also use the table and take the closest value). The sample t statistic value is calculated as follows:
Use Data Analysis in Excel to determine the mean and the sample standard deviation (2.505 and 1.5071 respectively). Then calculate the sample t statistic: .
Now we can make the following inference:
Since 4.6565 > 1.2858, there is sufficient evidence to support the hypothesis that the mean number of mutual funds is more than two at 10% significance. We can also use Data Analysis Plus. Select the option t-test: Mean.
b.The commission is paid erroneously if it is believed (based on the sample) that the mean is greater than 2 (that is, H0 is rejected) while in fact it is not (H0 is true). We need to calculate the probability of performing this (type I) error, which in fact is the p value when the sample statistic is t = 4.6565. From Data Analysis Plus the p value is ‘0’.
c.The commission is erroneously denied if there is insufficient evidence to support the hypothesis that >2 when in fact = 2.5. This is a type II error, and we are required to find its probability.
The two hypotheses are:
H0: = 2
H1: = 2.5.
The rejection region is, so the rejection region critical value in terms of the sample mean is:
3.a.We need to construct a confidence interval for the mean.
Note: 1- = .98, so /2 = .01. Thus we use t.01 in calculating the confidence interval.
b.The promotion is successful if the mean loan is greater than 60,500.
H0: = 60,500
H1: 60,500
Run a right-hand tail t-test. The rejection region: t t,n-1 = t.01, 170 = 2.3485 (see Data Analysis Plus).
The t statistic is: . Since .47 < 2.3485 there is insufficient evidence to reject the null hypothesis. At 1% significance level we cannot infer that the mean loan is more than $60,500. The promotion was not successful.
c.Let us define a success as: A customer borrows $65,000 or more. Thus, p is the probability that a customer borrows $65,000 or more.
H0: p = .45
H1: p > .45
Checking conditions: np = 171(.45) > 5; n(1- p) = 171(.55) > 5. So, the normal distribution can be used. The rejection region is Z > ZIf = .05 the rejection region is Z > 1.645. The Z statistic is calculated as follows:
.
[The sample proportion of 0.4912 was calculated by Data Analysis Plus].
Test conclusion: Since 1.0837 1.645, there is insufficient evidence to infer that more than 45% of the loans are for an amount of $65,000 or more. The bonus should not be paid.
Loans
Sample Proportion / 0.4912
Observations / 171
Hypothesized Proportion / 0.45
z Stat / 1.0837
P(Z<=z) one-tail / 0.1393
z Critical one-tail / 1.6449
P(Z<=z) two-tail / 0.2786
z Critical two-tail / 1.96
The Data Analysis Plus printout:
We can perform the test based on the
p-value. The p value for this test is .1393.
Since we selected a significance level of
= .05, the p value is greater than alpha. Again, we conclude that there is not enough
evidence to favor H1, or, there is
insufficient evidence to believe that the proportion is greater than .45.
- The largest sample needed for the a margin of error of W= .05 and a 90% confidence level can be found by
Taking the current sample proportion is an initial estimate, the sample size needed is