Section 7 3 Hypothesis Testing for the Mean

Section 7 -3 Hypothesis Testing for the Mean (Small Samples) Objective: SWBAT How to find critical values in a t- distribution. How to use the t-test to test a mean How to use technology to find P- values and use them with a t-test to test a mean

Critical values in a t- distribution • In section 7. 2 you learned how to perform a hypothesis test for a population mean when the sample size was 30 or more. In real life however it is often impractical to collect samples of size 30 or more. However if the population has a normal or nearly normal distribution you can still test the population mean ц. To do so you use the t-sampling distribution with n-1 degrees of freedom.

Guidelines Finding critical values in a t- distribution 1. Identify the level of significance α. 2. Identify the degrees of freedom d. f. = n-1 3. Find the critical values Using the t- table with n-1 Degrees of freedom. If the hypothesis test is a. Left tailed use One tail α column with a negative sign b. right tailed use One tail α column with a positive sign c. Two tailed use Two tails α column with a negative sign and a positive sign.

Example 1 Finding critical values for t Find the critical value to for a left tailed test given α=. 05 and n=21 Solution: The degrees of freedom are : d. f. = n-1 = 21 – 1 =20 Find the critical value in the. 05 column at the d. f = 20 the critical value is negative. So to = - 1. 725

Try it yourself Find a critical value for a left tailed test with α =. 01 and n = 14 a. Find the t- value in the t- table use d. f. = 13 and α =. 01 in the One tail α column. b. Use a negative sign.

Example 2 Finding the critical values for t Find the critical value to for a right tailed test with α =. 01 and n = 17 Solution : the degrees of freedom are: d. f. = n -1 = 17 – 1 = 16 Use the t- table we can use α =. 01 and d. f. =16 So to = 2. 583 ( the value is positive because it is for a right tailed test)

Try it yourself Find the critical value for a right tailed test with α =. 05 and n =9 a. Find the t value in the t- table using d. f = 8, And in the One tailed column

Example 3 Finding critical values for t Find the critical values for to and – to fro a two tailed test with α =. 05 and n = 26 Solution : the degrees of freedom are: n-1 = 26 – 1 = 25 because this is a two tailed test one value is positive and one is negative. So to = 2. . 060 and - to = -2. 060

Try it yourself Find the critical values ± to for a two tailed test With α =. 01 and n = 16 a. Find the t value in the t- table using d. f. =15 and α =. 01 in the two tail α column.

The t-test for a mean ц (n < 30, σ unknown) t = (sample mean – Hypothesized mean) Standard error The degrees of freedom are d. f. = n - 1

Guidelines Using the t test for a mean ц (Small Sample) 1. Stat the claim mathematically identify The null and alternative hypothesis 2. 3. Idetify the level of significance Identify the degrees of freedom and sketch the sampling distribution 4. Determine any critical values 5. Determine any rejection regions 6. Find the standardized test statistic. 7. Make a deciasion to reject or fail to reject the null hypothesis. Ho and Ha Identify α d. f. = n - 1 Use the t – table If t is in the rejection region Reject Ho Otherwise fail to reject Ho. 8. Interpret the decision in the context of the Original claim.

Remember that when you make a decision the possibility of a type I or a type II error exists. We will later discuss how to use a P value for a t - test for a mean ц (small sample).

Example 4 A used car dealer says that the mean price of a 2002 Ford F-150 Super Cab is at least $18, 800. You suspect that this claim is incorrect and find that a random sample of 14 similar has a mean price of $18000 and a standard deviation of $1250. Is there enough evidence to reject the dealer’s claim at α =. 05 Assume the population is normally distributed.

Solution: The claim is that the mean price is at least $18, 800. So the null and alternative hypothesis are Ho: ц ≥ $18800 and H a < 18800 The test is a left tailed test the level of significance is α =. 05 and there are d. f. = 14 -1=13 degrees of freedom. So the critical value is to =-1. 771 the rejection region is t < - 1. 771 = ≈ -2. 39

Interpretation Because t is in the rejection region you decide to reject the null hypothesis. There is enough evidence at the 5% level of significance to reject the claim that the mean price of a 2002 For F-150 Super Cab is at least $18800.

Try it yourself An industrial company claims that the p. H level of the water in the nearby river is 6. 8. You randomly select 19 water samples and measure the p. H of each. The sample mean and the standard deviation are 6. 7 and. 24 respectively. Is there enough evidence to reject the company’s claim at α =. 05? Assume the population is normally distributed.

The t Sampling Distribution Find the critical value t 0 for a left-tailed test given 18. = 0. 01 and n = d. f. = 18 – 1 = 17 Area in left tail t 0 = – 2. 567 t 0 Find the critical values –t 0 and t 0 for a two-tailed test given = 0. 05 and n = 11. –t 0 = – 2. 228 and t 0 = 2. 228 d. f. = 11 – 1 = 10 t 0

Testing –Small Sample A university says the mean number of classroom hours per week for full-time faculty is 11. 0. A random sample of the number of classroom hours for full-time faculty for one week is listed below. You work for a student organization and are asked to test this claim. At = 0. 01, do you have enough evidence to reject the university’s claim? 11. 8 8. 6 12. 6 7. 9 6. 4 10. 4 13. 6 9. 1 1. Write the null and alternative hypothesis 2. State the level of significance = 0. 01 3. Determine the sampling distribution Since the sample size is 8, the sampling distribution is a t -distribution with 8 – 1 = 7 d. f.

Since Ha contains the ≠ symbol, this is a two-tail test. 4. Find the critical values. 5. Find the rejection region. –t 0 – 3. 499 t 0 3. 499 6. Find the test statistic and standardize it n=8 = 10. 050 s = 2. 485 7. Make your decision. t = – 1. 08 does not fall in the rejection region, so fail to reject H 0 at = 0. 01 8. Interpret your decision. There is not enough evidence to reject the university’s claim that faculty spend a mean of 11 classroom hours.

Minitab Solution Enter the data in C 1, ‘Hours’. Choose t-test in the STAT menu. T-Test of the Mean Test of = 11. 000 vs not = 11. 000 Variable N Mean St. Dev SE Mean T Hours 8 0. 050 2. 485 0. 879 – 1. 08 Minitab reports the t-statistic and the P-value. Since the P-value is greater than the level of significance (0. 32 > 0. 01), fail to reject the null hypothesis at the 0. 01 level of significance. P 0. 32

Homework 1 -25 odd. Pg. 370 -371 Day 2: 26 -34 all pgs. 371 -373