A random sample of 225 yielded a simple proportion of 0.46. Construct a 95% conf

Question

A random sample of 225 yielded a simple proportion of 0.46. Construct a 95% confidence interval for the population proportion. A random sample of 100 observations from a population with a population standard deviation 6 yielded a sample mean of 110.Test the hypothesis that the mean is greater than 100 using alpha = 0.05. The reputations of many businesses can be severely damaged by shipments of manufactured items that contain a large percentage of defectives. Suppose 300 batteries are randomly selected, tested, and found that 10 are defective. Does this provide sufficient evidence that the fraction of defectives in the entire is less than 0.05? Use alpha = 0.01. In recent years, the US and Japan have engaged in intense negotiations regarding restrictions on trade between the two countries. An economist decides to test the hypothesis that higher retail prices are being charged for Japanese automobiles in Japan than in the US., She randomly obtained a sample of 50 retail sales in the US and 30 retail sales in Japan over the same time period and for the same automobile A summary table is below: Test the hypothesis at a0.05 level of significance and assume the population variances are equal. A manufacturing operation consists of a single-machine-tool system that produces an average of 15.5 transformer parts every hour. The production of transformers is assumed to be normally distributed. The system was monitored and observed the number of parts produced in each of 17 randomly selected one-hour periods. The mean and standard deviation of the 17 production runs arc 15.42 and 0.16, respectively. Does this provide evidence that the true mean number of parts produced every hour Jitter's from 15.5? Test at alpha = 0.05. According to a CCH Unscheduled Absence Survey, 9% of small businesses use telecommunicating of workers in an effort to reduce unscheduled absenteeism. This proportion compares to 6% for all businesses. Is there really a significant difference between small business and all business in this issue? Use the data and an alpha of 0.1 to test this question. Assume that here were 780 small businesses and 915 other businesses in this survey.

Explanation / Answer

Problem 1

We are given

Sample size = n = 225

Sample proportion = P = 0.46

Confidence level = 95%

Critical Z value = 1.96

Standard error of the proportion = sqrt(P(1 – P)/n) = sqrt(0.46(1 – 0.46)/225) = 0.0332

Margin of error = Critical Z * Standard error = 1.96*0.0332 = 0.0651

Lower limit = P – Margin of error = 0.46 – 0.0651 = 0.3949

Upper limit = P + Margin of error = 0.46 + 0.0651 = 0.5251

Confidence Interval = (0.3949, 0.5251)

Problem 2

H0: µ 100 V/s Ha: µ > 100

Z = (Xbar - µ)/[/sqrt(n)]

We are given

Xbar = 110

Population mean = µ = 100

Population SD = = 6

Sample size = n = 100

Level of significance = = 0.05

Critical value = 1.6449

Z = (110 – 100)/[6/sqrt(100)] = 10/[6/10] = 10/0.6

Z = 16.6667

P-value = 0.00

= 0.05

P-value < = 0.05

So, we reject the null hypothesis.

Problem 3

H0: p 0.05 V/s Ha: p < 0.05

Z = (P – p)/sqrt(pq/n)

Where, P = 10/300 = 0.03333, p = 0.05, q = 1 – p = 1 – 0.05 = 0.95, n = 300, = 0.01

Z = (0.03333 – 0.05)/sqrt(0.05*0.95/300) = -1.3245

Lower critical value = -2.3263

P-value = 0.0927

= 0.01

P-value > = 0.01

So, we do not reject the null hypothesis.

Problem 4

H0: µ1 = µ2 V/s Ha: µ1 > µ2 (One tailed – upper tailed test)

Test statistic formula

t = (X1bar – X2bar) / sqrt[(S1^2/N1)+(S2^2/N2)]

t = (17.243 – 16.545) / sqrt[(1.843^2/30)+(1.989^2/50)]

t = 1.5612

Degrees of freedom = 30 + 50 – 2 = 78

Critical value = 1.6646

P-value = 0.0613

= 0.05

P-value > = 0.05

So, we do not reject the null hypothesis

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