Question

1) For the following data values below, construct a 90% confidence interval if the sample mean is known to be 0.719 and the standard deviation is 0.366. (Round to the nearest thousandth) (Type your answer in using parentheses! Use a comma when inputing your answers! Do not type any unnecessary spaces! List your answers in ascending order!) for example: (0.45,0.78) 0.56, 0.75, 0.10, 0.95, 1.25, 0.54, 0.88

2) For the following data values below, construct a 98% confidence interval if the sample mean is known to be 9.808 and the population standard deviation is 5.013. (Round to the nearest thousandth) (Type your answer in using parentheses! Use a comma when inputing your answers! Do not type any unnecessary spaces! List your answers in ascending order!) for example: (0.45,0.78) 6.6, 2.2, 18.5, 7.0, 13.7, 5.4, 5.3, 5.9, 4.7, 14.5 2.0, 14.8, 8.1, 18.6, 4.5, 17.7, 15.9, 15.1, 8.6, 5.2 15.3, 5.6, 10.0, 8.2, 8.3, 9.9, 13.7, 8.5, 8.2, 7.9 17.2, 6.1, 13.7, 5.7, 6.0, 17.3, 4.2, 14.7, 15.2, 3.3 3.2, 9.1, 8.0, 18.9, 14.2, 5.1, 5.7, 16.4, 10.1, 6.4

3)In a randomized controlled trial in Kenya, insecticide treated bednets were tested as a way to reduce malaria. Among 343 infants using bednets, 15 developed malaria. Among 294 infants not using bednets, 27 developed malaria. Want to use a 0.01 significance level to test the claim that the incidence of malaria is lower for infants using bednets. Find the test statistic. (Round to the nearest hundredth) (ONLY TYPE IN THE NUMBER!)

4)State the conclusion whether or not to REJECT or FAIL TO REJECT the null hypotheses.(Check your spelling!) The original claim: The percentage of of M&Ms is greater than 5%. The hypothesis test results in a p-value of 0.0010. a=0.05

Answer 1

1)

Since, we do not know the population standard deviation , we will use t statistic to construct 90% confidence interval.

Degree of freedom = n - 1 = 7 - 1 = 6

Critical value of t at 90% confidence interval and df = 6 is 1.94

Margin of error = t * standard deviation = 1.94 * 0.366 = 0.71

90% confidence interval of mean is,

(0.719 - 0.71, 0.719 + 0.71)

(0.01, 1.43)

2)

Since, we know the population standard deviation , we will use z statistic to construct 98% confidence interval.

Critical value of z at 98% confidence interval is 2.33

Margin of error = t * standard deviation = 2.33 * 5.013 = 11.680

98% confidence interval of mean is,

(9.808 - 11.68, 9.808 + 11.68)

(-1.872, 21.488)

3)

H0: Incidence of malaria for infants using bednets is equal to the infants not using bednets.

H1: Incidence of malaria for infants using bednets is lower than the infants not using bednets.

p1 = 15 / 343 = 0.0437

p2 = 27 / 294 = 0.0918

Pooled proportion, p = (x1 + x2) / (p1 + p2) = (15 + 27) / (343 + 294) = 0.0659

Standard error of difference in proportions =

= 0.0197

Test statistic, z = (p1 - p2) / Std error

= (0.0437 - 0.0918) / 0.0197

= -2.44

4)

P-value = P[z < -2.44] = 0.0073

Since p-value is less than 0.01 significance level, we reject null hypothesis H0 and conclude that there is significant evidence that incidence of malaria for infants using bednets is lower than the infants not using bednets.

The original claim: Incidence of malaria for infants using bednets is equal to the infants not using bednets.

The hypothesis test results in a p-value of 0.0073. a=0.01