Its due tonight. Thank you. A sociologist is studying the age of the population in Blue Valley.

Please help, i am struggling to understand. Its due tonight. Thank you.

19%13%32%24%12%

Under 20 20 - 35 36 - 50 51 - 65 Over 65

29 26 67 66 22

(i) Give the value of the level of significance. 

State the null and alternate hypotheses. 

H0: The distributions for the population 10 years ago and the population today are the same.

H1: The distributions for the population 10 years ago and the population today are different.

H0: Time ten years ago and today are independent.

H1: Time ten years ago and today are not independent.

H0: Ages under 20 years old, 20- to 35-year-old, between 36 and 50, between 51 and 65, and over 65 are independent.

H1: Ages under 20 years old, 20- to 35-year-old, between 36 and 50, between 51 and 65, and over 65 are not independent.

H0: The population 10 years ago and the population today are independent.

H1: The population 10 years ago and the population today are not independent.

(ii) Find the sample test statistic. (Round your answer to two decimal places.) 

(iii) Find or estimate the P-value of the sample test statistic. 

P-value > 0.100

0.050 < P-value < 0.100

0.025 < P-value < 0.050

0.010 < P-value < 0.025

0.005 < P-value < 0.010

P-value < 0.005

(iv) Conclude the test. 

Since the P-value ≥ α, we do not reject the null hypothesis.

Since the P-value < α, we do not reject the null hypothesis. 

Since the P-value < α, we reject the null hypothesis.

Since the P-value ≥ α, we reject the null hypothesis.

(v) Interpret the conclusion in the context of the application. 

At the 1% level of significance, there is insufficient evidence to claim that the age distribution of the population of Blue Valley has changed.

At the 1% level of significance, there is sufficient evidence to claim that the age distribution of the population of Blue Valley has changed.

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21s2

27s2

Classify the problem as being a Chi-square test of independence or homogeneity, Chi-square goodness-of-fit, Chi-square for testing or estimating σ2 or σ, F test for two variances, One-way ANOVA, or Two-way ANOVA, then perform the following. 

One-way ANOVA

Two-way ANOVA

Chi-square test of independence

F test for two variances

Chi-square test of homogeneity

Chi-square goodness-of-fit

Chi-square for testing or estimating σ2 or σ

(i) Give the value of the level of significance. 

State the null and alternate hypotheses. 

H0: σ12 = σ22; H1: σ12 > σ22

H0: σ12 = σ22; H1: σ12 ≠ σ22

H0: σ12 < σ22; H1: σ12 = σ22

H0: σ12 = σ22; H1: σ12 < σ22

(ii) Find the sample test statistic. (Round your answer to two decimal places.) 

(iii) Find the P-value of the sample test statistic. 

P-value > 0.200

0.100 < P-value < 0.200

0.050 < P-value < 0.100

0.020 < P-value < 0.050

0.002 < P-value < 0.020

P-value < 0.002

(iv) Conclude the test. 

Since the P-value is greater than or equal to the level of significance α = 0.05, we fail to reject the null hypothesis.

Since the P-value is less than the level of significance α = 0.05, we reject the null hypothesis.

Since the P-value is less than the level of significance α = 0.05, we fail to reject the null hypothesis.

Since the P-value is greater than or equal to the level of significance α = 0.05, we reject the null hypothesis.

(v) Interpret the conclusion in the context of the application. 

At the 5% level of significance, there is insufficient evidence to show that the variance for the new manufacturing process is different.

At the 5% level of significance, there is sufficient evidence to show that the variance for the new manufacturing process is not different.

At the 5% level of significance, there is insufficient evidence to show that the variance for the new manufacturing process is not different.

At the 5% level of significance, there is sufficient evidence to show that the variance for the new manufacturing process is different.