Testing Means with small samples Critical Value from the T-table –Use the one-tail heading for...

Testing Means with small samples

• Critical Value from the T-table– Use the one-tail heading for left or right tail

• Make negative for left-tail

– Use the two-tails heading for two tailed test

– Remember degrees of freedom

• For Example1. H0: μ ≥ 25, n = 18, Degree of confidence = 99%

2. H0: μ = 110, n = 25, Degree of confidence = 95%

3. H0: μ ≤ 1.5, n = 12, Degree of confidence = 90%

4. H0: μ = 74, n = 28, Degree of confidence = 99.9%

Live example/Your Turn

• What would you guess to be average age of South students’ cars?

– Let’s do a survey!!

• What would you guess to be minimum number of colleges that average South student’s applied to?

– Let’s take a sample!!

Worksheet

Sample mean = n = s =

Original claim •Label null and alternative hypothesisOpposite claim

Degree of confidence

Test statistics

Critical region•Two-tailed (H0 =)

•Left tailed (H0 )

•Right tailed (H0 )

Critical value

Reject or accept?

Testing Hypothesis with Small Samples

• We use the T (Student) table (A-3) to find the critical value– We need to know the degrees of freedom, the significance level

(alpha), and the number of tails

• Calculator:– [STAT]

– TESTS

– 1: T-Test…

• μ0 is the benchmark.

• X-bar is the mean

• Sx is the standard deviation

• n is the sample size

• μ select the format of H1

Homework: Test the claims

1. Claim: Student population has a mean GPA of 2.0. A sample of 24 students has a mean is 2.35 and a standard deviation is 0.70. Use a 95% degree of confidence

2. Claim: An SAT prep class produces scores above 1700. A sample of 15 students has a mean is 1685 and a standard deviation is 170. Use a 99% degree of confidence

3. Claim: The average college student needs at least 5 years to get a degree. A sample of 20 students has a mean of 5.15 years and the standard deviation is 1.68. Use a 90% degree of confidence

4. The following list contains the repair costs for five BMW cars used in a controlled crash test: 797 571 904 1147 418. Use this sample to text the claim that BMW’s repair costs are under $1000.

5. Using a sample of 25 adults whose mean body temperature was 98.24 (standard deviation = 0.56), Test the claim that the mean body temperature for the population is 98.6

Homework #1

Sample mean = 2.35 n = 24 s = 0.07

Original claim µ = 2.0 (H0) Label null and alternative hypothesis

Opposite claim µ ≠ 2.0 (H1)

Degree of confidence 95%

Test statistics t = 2.45




Critical value

Reject or accept?

p-value = 0.022 < 0.05 Reject null, reject original claim

Homework #2

Sample mean = 1685 n = 15 s = 170

Original claim µ > 1700 (H0) Label null and alternative hypothesis

Opposite claim µ ≤ 1700 (H1)


Test statistics t = -0.342




Critical value

Reject or accept?

p-value = 0.631 > 0.01 FRT null, FTR original claim

Homework #3

Sample mean = 5.15 n = 20 s = 1.68

Original claim µ ≥ 5(H0) Label null and alternative hypothesis

Opposite claim µ < 5 (H1)


Test statistics t = 0.399




Critical value

Reject or accept?

p-value = 0.653 > 0.1 FRT null, FRT original claim

Homework #4

Sample mean = 588 n = 5 s = 424

Original claim µ < 1000 (H1) Label null and alternative hypothesis

Opposite claim µ ≥ 1000 (H0)






Critical value

Reject or accept?

p-value = 0.048 < 0.05 Reject null, accept original claim

Homework #5

Sample mean = 98.24 n = 25 s = 0.56

Original claim µ = 98.6 (H0) Label null and alternative hypothesis

Opposite claim µ ≠ 98.6 (H1)






Critical value

Reject or accept?

p-value = 0.0037 < 0.05 Reject null, reject original claim

Testing Means with small samples Critical Value from the T-table –Use the one-tail heading for...

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