Tests about a

Population

MeanChapter 9, lesson 3

Warm Up

Please find the descriptive statistics and

any data displays for the data so that you

are able to describe the distribution:

44 49 45 51 49 53

49 44 47 50 46 48

Less music? A classic rock radio station claims to play an

average of 50 minutes of music every hour. However, I seems that every time you turn to this station, there is a commercial playing. To investigate the station’s claim, you randomly select 12 different hours during the next week and record what the radio station plays in each of the 12 hours. Here are the number of minutes of music played during each of these hours.

Check the conditions for carrying out a significance test of the station’s claim that it plays an average of 50 minutes of music per hour.

44 49 45 51 49 53

49 44 47 50 46 48

Conditions for performing a

significance test about a mean

1. Random?

SRS?

2. 10%:

Must check that EQUATION if sampling without replacement,

3. Normal/large counts:

If the sample size is large (n ≥ 30), or has a Normal distribution, you’re good to go.

If the sample size is small (n < 30), it must be confirmed that the sample data is not strongly skewed or have outliers.

Calculations: Test Value

𝑡 =ҧ𝑥 − 𝜇𝑜

ൗ𝑠𝑥

𝑛

P-Value P-Value for t-distributions are described as an interval, and this

interval is compared to α.

1. Once the t statistic has been calculated, bust out the t table to look for your p-value.

1. Look in the row of your degrees of freedom; still round down if the exact df aren’t there.

1. The t-statistic you computed won’t be there, but find between what two columns it is between.

1. At the top, identify the tail probabilities p that correspond to these t statistics.

2. Express the P-Value as an interval between these two; if two-tail, double the tail areas.

Back to the example

Step 3: DO

1. Calculate the t-statistic

t = -2.59

2. Calculate the P-Value

with df = 11

0.01 < P-Value < 0.02

Example 2: Calc P-value

a. Find the P-value for a test of Ho: μ = 10 vs. HA:

μ > 10 that uses a sample of size 75 and has

a test statistic of t = 2.33.

• df = 74 so round down to 60

• t = 2.33 is between 2.099 and 2.390

• 0.01 < P-value < 0.02

Example 2: Calc P-value

b. Find the P-value for a test of Ho: μ = 300 vs.

HA: μ ≠ 300 that uses a sample of size 10 and

has a test statistic of t = -0.51.

• df = 9

• Area under the t curve to the right of 0.51 is

greater than 0.25.

• Because this is a two-sided test, the P-value

must be at least 0.50.

Example 3: Short Subs

You notice that the lengths of the “6-inch

sub” sandwiches they get at their favorite

restaurant seemed shorter than the

advertised length. To investigate, you

randomly select 24 different times during

the next month and ordered a “6-inch” sub.

Here are the actual lengths of each of the

24 sandwiches (in inches):

Short Subs

(just a suggestion: work with a

neighbor: have one person read off

data, the other enter in the calculator)

a. Do these data provide convincing

evidence at the α=0.10 level that the

sandwiches at this restaurant are shorter

than advertised, on average?

4.50 4.75 4.75 5.00 5.00 5.00 5.50 5.50

5.50. 5.50 5.50 5.50 5.75 5.75 5.75 6.00

6.00 6.00 6.00 6.00 6.50 6.75 6.75 7.00

Step 1

State: Solution

State:

we want to test the following hypotheses

at α = 0.10 significance level:

μ= the true mean length of “6-inch” subs

from this restaurant

Ho: μ = 6, HA: μ < 6

Step 2:

Plan: Solution

We will make a one-sample t test for μ.

Random?

Yes, the times were randomly selected

10%?

I should hope that the restaurant makes more

than 240 sandwiches per month.

Normality/Large Numbers?

n ≤ 30 so must check: Based on the distribution,

there are no outliers or real skewedness.

Step 3:

Do

Test statistic:

t = -2.38

P-value:

df = 23

0.01 < P-value < 0.02

Step 4:

Conclude

Because the P-value, which is between

0.01 and 0.02, is less than α = 0.10, we

reject the null hypothesis.

There is convincing evidence that the true

mean length of “6-inch” subs at this

restaurant is less than 6 inches.

Part b.

b. Given your conclusion in part (a), which

kind of mistake- a Type I or Type II error-

could you have made? Explain what this

mistake would mean in context?

Part b.

b. Given your conclusion in part (a), which kind of mistake- a Type I or Type II error-could you have made? Explain what this mistake would mean in context?

Because we rejected the null hypothesis, it is possible that we made a Type I error. In other words, it is possible that we found convincing evidence that the mean length was less than 6 inches when in reality the mean length is 6 inches.

Example 4

Don’t break the iceIn the children’s game Don’t Break the Ice, small plastic ice cubes are squeezed into a square frame. Each child takes turns tapping out a cube of “ice” with a plastic hammer hoping that the remaining cubes don’t collapse. For the game to work correctly, the cubes must be big enough so that they hold each other in place in the plastic frame but not so big that they are too difficult to tap out. The machine that produces the plastic ice cubes is designed to make cubes that are 29.5 mm wide, but the actual width varies a little. To make sure the machine is working well, each hour a supervisor inspects a random sample of 50 cubes every hour and measures their width. During one hour, a sample had a mean of 29.4846 mm, a standard deviation of 0.090 mm, and a standard error of 0.013 mm.

Problem:

a. Do these data give convincing evidence that the mean width of cubes produced this hour is not 29.5 mm?

Step 1

State: Solution

State:

we want to test the following hypotheses

at α = 0.05 significance level:

μ= the true mean width of plastic cubes

Ho: μ = 29.5, HA: μ ≠ 29.5

Step 2:

Plan: Solution

We will make a one-sample t test for μ.

Random?

Yes, a random sample was taken during an hour. 10%?

I would assume that the machine produces more than 500 cubes per hour.

Normality/Large Numbers?

n ≥ 30 so because of the central limit theorem, we can be sure that the sampling distribution is normal.

Step 3:

Do

Test statistic:

t = -1.21

P-value:

df = 40

0.10 < P-value < 0.15

2(0.10) < 2(P-value) < 2(0.15)

0.20 < P-value < 0.30

Step 4:

Conclude

Because the P-value, between 0.20 and

0.30 is greater thanα = 0.05, we fail to

reject the null hypothesis.

There is not convincing evidence that the

true width of the plastic ice cubes

produced this hour is different from 29.5

mm.

Don’t Break the Ice b. Interpret the standard deviation and the

standard error provided by the computer output.

Standard deviation: the average difference (0.090 mm) of individual cubes from the mean width (29.4846 mm).

Standard Error: the average difference (0.013) the mean of a sample of 50 from the true mean.

Confidence Intervals

There is a relationship between the decision made in a significance test and a confidence interval.

If done correctly, the hypothesized null value should be in an interval of the same level if the null was failed to be rejected.

The hypothesized null value should not be in an interval of the same level if the null was rejected.

Don’t Break the Ice

a. Construct the 95% interval for the true

mean width of plastic ice cubes

produced this hour.

b. Would you make the same conclusion

with the confidence interval as you did

the significance test? Justify your

conclusion.

c. Interpret the confidence level.

Don’t Break the Ice

a. Construct the 95% interval for the true mean width of plastic ice cubes produced this hour.

𝟐𝟗. 𝟒𝟓𝟗𝟎𝒎𝒎 < 𝝁 < 𝟐𝟗. 𝟓𝟏𝟎𝟐𝒎𝒎

a. Would you make the same conclusion with the confidence interval as you did the significance test? Justify your conclusion.

b. Interpret the confidence level.