Question

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5. (1 pts) You are conducting a study to see if the probability of a true negative on a test for a certain cancer is significantly less than 0.59. Thus you are performing a test statistic z -1.835. Find the p-value accurate to 4 decimal places. Label the standard normal curve and shade the appropriate area. -tailed test. Your sample data produce the p-value= 6. (6 pts) Test the claim that the proportion of people who own cats is significantly different than 30% at the 0.02 significance level. The null and alternative hypothesis would be: Ho H1 Based on a sample of 500 people, 135 owned cats The test statistic is: (to 2 decimals) The p-value is: (to 2 decimals) Based on this we: Reject the null hypothesis Fail to reject the null hypothesis Conclusion:

Answer 1

5. You are performing a left(or one) tailed test.

z = -1.835

p-value = NORM.S.DIST(-1.835, 1) = 0.0333

p-value 0.0333 2 3 -1

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6. n = 500, x = 135

p̄ = x/n = 0.27

Null and Alternative hypothesis:

Ho : p = 0.30

H1 : p ≠ 0.30

Test statistic:   

z =(p̄ -p)/√(p*(1-p)/n) = -1.4639  

p-value = 2*(1-NORM.S.DIST(ABS(-1.4639), 1)) = 0.1432

Decision:   

p-value > α, fail to reject the null hypothesis.

Conclusion:

There is not enough evidence to conclude that the proportion of people who own cats is significantly different than 30% at 0.02 significance level.

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9. x̅ = 68.5, s = 8.7, n = 28

α = 0.10

Null and Alternative hypothesis:

Ho : µ = 62.9

Ha : µ > 62.9

Test statistic:

t = (x̅- µ)/(s/√n) = (68.5 - 62.9)/(8.7/√28) = 3.406

df = n-1 = 27

p-value = T.DIST.RT(3.406, 27) = 0.0010

Decision:

p-value < α, Reject the null hypothesis

Conclusion:

There is enough evidence to conclude that the population mean is greater than 62.9 at 0.10 significance level.

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10. Let, proportion, p = 0.50

Margin of error, E = 5% = 0.05

Confidence Level, CL = 0.90

Significance level, α = 1 - CL = 0.10

Critical value, z = NORM.S.INV(0.10/2) = 1.6449

Sample size, n = (z² * p * (1-p)) / E² = (1.6449² * 0.5 * 0.5)/ 0.05²

= 270.5543 = 271