WARM – UP ( just do “a.” )

The Gallup Poll asked a random sample of 1785 adults, “Did you attend church or synagogue in the last 7 days?” Of the respondents, 750 said “Yes.”

a.)Find the 99% Confidence Interval for the % of all adults who attended church or synagogue during the past week. (Follow all steps)

b.)Do the results provide good evidence that less than half of the population attend church or synagogue? Give supporting evidence? (Follow all steps)

ˆ ˆ1ˆ

p pp z

n

.420 1 .420

0.420 (2.576)1785

We can be 99% confident that the true proportion of We can be 99% confident that the true proportion of adults who attended church or synagogue in the past 7 adults who attended church or synagogue in the past 7 days is between 39.0% and 45.0%.days is between 39.0% and 45.0%.

1.1. SRS – Stated SRS – Stated 2.2. Appr. Normal: Appr. Normal: 1785(1785(0.4200.420) ) ≥ 10 1785(1 – ≥ 10 1785(1 – 0.4200.420) ≥ 10) ≥ 103.3. Population of Adults Population of Adults ≥ 10(1785)≥ 10(1785)

One ProportionOne ProportionZ– Conf. Int.Z– Conf. Int.

a.) Find the 99% Confidence Interval for the % of all adults who attended church or synagogue during the past

week. (Follow all steps)

The Gallup Poll asked a random sample of 1785 adults, “Did you attend church or synagogue in the last 7 days?” Of the respondents, 750 said “Yes.”

P-value ≤ 0.05P-value ≤ 0.05-Reject H0

-Evidence supporting Ha

P-value > 0.05P-value > 0.05- Fail to Reject H0

- No Evidence supporting Ha

Decision Decision ConclusionConclusion

Chapter 21 – More about Tests

b.) Of 1785, 750 said “Yes.” Do the results provide good evidence that less than half of the population attend church or synagogue? Give supporting evidence? (Follow all steps)

6.7457z ( 99, 6.7457)P Value Normalcdf E

Since the P-Value is less than Since the P-Value is less than αα = 0.05 we REJECT H = 0.05 we REJECT H00. There is . There is

sufficient evidence that The % of adults attending church or sufficient evidence that The % of adults attending church or synagogue is less than 50%.synagogue is less than 50%.

p = The true proportion of adults that attend church or p = The true proportion of adults that attend church or synagogue in the past 7 days. synagogue in the past 7 days.

HH00: : p = p = 0.500.50

HHaa: : p p < 0.50< 0.50

0

0 0

ˆ

1

p pz

p p

n

0.42 0.50

0.5 1 0.5

1785

z

One Proportion One Proportion

z – Testz – Test

1.1. SRS: Stated SRS: Stated 2.2. Appr. Normal: Appr. Normal: 1785(1785(0.500.50) ) ≥ 10 1785(1 – ≥ 10 1785(1 – 0.500.50) ≥ 10) ≥ 103.3. Population of Adults Population of Adults ≥ 10(1785)≥ 10(1785)

1.

2. 3.

4.

5.

6.

0

p = The true proportion of Hispanics called for jury duty in the county.

H0 : p = 0.19 HA : p < 0.19

p=The true proportion of Hispanics called for jury in the county.H0 : p = 0.19 HA : p < 0.19

0

0 0

ˆ

1

p pz

p p

n

0.125 0.19

0.19 1 0.19

72

z

One Proportion One Proportion

z – Testz – Test

SRS: potential jurors are called randomly from all of the residents.10% condition: Population of jurors in the county > 10(72)Success/Failure condition: np= (72)(0.19) = 13.68 and n(1 – p) = (72)(0.81) = 58.32 are both > 10, so the sample is large enough.The conditions have been satisfied, so a Normal model can be used to model the sampling distribution of the proportion,

z 1.41 0.0793P Value ( 99, 1.41)normalcdf E Since the P-value = 0.0793 is somewhat high p>0.05, we fail to reject the null hypothesis. There is NO convincing evidence that Hispanics are underrepresented in the jury selection system.

• What can you say if the P-value does not fall below ? – You should say that “The data have failed to provide

sufficient evidence to reject the null hypothesis.” – NEVER “accept the null hypothesis”. ONLY “Fail to

Reject the null hypothesis.”

Chapter 21 – More about Tests

• Common alpha levels are 0.10, 0.05, and 0.01. – When we reject the null hypothesis, we say that the

test is “significant at that level.”

• The alpha level, α =0.05,is also called the significance significance level. level.

Homework- Page 491: 1,2,5a,6a, 8-10

H0 : p = 0.40 (p=The proportion of female executives.)HA : p < 0.40

H0 : p = 0.40 (The proportion of female executives.)HA : p < 0.40

0

0 0

ˆ

1

p pz

p p

n

0.302 0.4

0.4 1 0.4

43

z

One Proportion One Proportion

z – Testz – Test

Randomization condition: Executives were not chosen randomly, 10% condition: Population of employees at the company > 10(43)Success/Failure condition: npnp= (43)(0.40) = 17.2 = (43)(0.40) = 17.2 and n(1 – p) n(1 – p) = (43)= (43)(0.60) = 25.8(0.60) = 25.8 are both >10, so the sample is large enough.The conditions have not all been satisfied, so a Normal model may or may not be valid, so we continue with Cautionso we continue with Caution.

Since the P-value = 0.0955 is high (p>0.05), we fail to reject the Hfail to reject the H00. There is little evidence to suggest proportion of female executives is any different from the overall proportion of 40% female employees at the company.

z 1.31 0.0955P Value

• When the alternative is two-sided, the critical value splits equally into two tails:

Confidence Intervals and Hypothesis Tests

A confidence interval with a confidence level of C% corresponds to a two-sided hypothesis test (The “≠” test) with an -level of 100 – C%. If the test “REJECTS H0”, then the sample statistic will NOT be in the interval

You can also reject using just the z-score. – Any z-score larger in magnitude than a particular critical value leads us to reject H0. – Any z-score smaller in magnitude than a particular

critical value leads us to fail to reject H0.

1-sided 2-sided

0.01 2.28 2.575

0.05 1.645 1.96

0.10 1.282 1.645

1.28

21.

645

2.28