Solve the problem. 3) A manufacturer finds that in a random sample of 100 of its

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Solve the problem. 3) A manufacturer finds that in a random sample of 100 of its CD players, 96% have no defects. The manufacturer wishes to make a claim about the percentage of nondefective CD players and is prepared to exaggerate. What is the highest rate of nondefective CD players that the manufacturer could claim under the following condition?

His claim would not be rejected at the 0.05 significance level if this sample data were used. Assume that a left-tailed hypothesis test would be performed.

4) The health of employees is monitored by periodically weighing them in. A sample of 54 employees has a mean weight of 183.9 lb. Assuming that is known to be 121.2 lb, use a 0.10 significance level to test the claim that the population mean of all such employees weights is less than 200 lb. Test the given claim. Use the P-value method or the traditional method as indicated. Identify the null hypothesis, alternative hypothesis, test statistic, critical value(s) or P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim.

5) A simple random sample of 15-year old boys from one cityis obtained and their weights (in pounds) are listed below. Use a 0.01 significance level to test the claim that these sample weights come from a population with a mean equal to 149 lb. Assume that the standard deviation of the weights of all 15-year old boys in the city is known to be 16.2 lb. Use the traditional method of testing hypotheses. 147 138 162 151 134 189 157 144 175 127 164 Assume that a simple random sample has been selected from a normally distributed population and test the given claim. Use either the traditional method or P-value method as indicated. Identify the null and alternative hypotheses, test statistic, critical value(s) or P-value (or range of P-values) as appropriate, and state the final conclusion that addresses the original claim.

6) A cereal company claims that the mean weight of the cereal in its packets is 14 oz. The weights (in ounces) of the cereal in a random sample of 8 of its cereal packets are listed below. 14.6 13.8 14.1 13.7 14.0 14.4 13.6 14.2 Test the claim at the 0.01 significance level. Assume that a simple random sample has been selected from a normally distributed population. Find the test statistic, P-value, critical value(s), and state the final conclusion.

7) Test the claim that the mean lifetime of car engines of a particular type is greater than 220,000 miles. Sample data are summarized as n = 23,x= 226,450 miles, and s = 11,500 miles. Use a significance level of = 0.01.

Find the critical value or values of 2 based on the given information.

8) H1: < 0.14 n = 23 = 0.10

Assume that the data has a normal distribution and the number of observations is greater than fifty. Find the critical z value used to test a null hypothesis.

9) = 0.08; H1 is µJ 3.24 Solve the problem.

10) A simple random sample of students is selected, and the students are asked how much time they spent preparing for a test.

The times (in hours) are as follows: 1.3 7.2 4.2 12.5 6.6 2.5 5.5 Based on these results, a confidence interval for the population mean is found to be µ= 5.7 ± 4.4. Find the degree of confidence.

Explanation / Answer

His claim would not be rejected at the 0.05 significance level if this sample data were used. Assume that a left-tailed hypothesis test would be performed.

Here we have given that,

n = 100

p^ = 96% = 0.96

alpha = 0.05

Now we have to find population proporiton (p) from the given information.

Here we have to test the hypothesis that,

H0 : p = p0 Vs H1 : p < p0

where p0 is the population proportion.

We have to find p0.

The test statistic follows Z-distribution.

Test statistic is,

Z = (p^ - p) / sqrt((p*q)/n)

For alpha = 0.05 , Zc = critical value of normal distribution at 0.05 level of significance = -1.645

Now we have to find p such that Zc = -1.645

Or (p^ - p) / sqrt((p*q)/n) = -1.645

(0.96 - p) / sqrt(p*q)/n = -1.645

Use trial and error method.

Here we see that for p = 0.982 we get test statistic = -1.6547

p = 0.982

For p = 0.982 all the conditions are met.

Here we have given that,

H0 : mu = 200 Vs H1 : mu < 200

where mu is population mean.

Assume alpha = level of significance = 0.10

GIven that,

Xbar = 183.9

sigma = 121.2

n = 54

Here population standard deviation is known therefore we use one sample z-test.

The test statistic is,

Z = (Xbar - mu) / (sigma/sqrt(n))

= (183.9 - 200) / (121.2/sqrt(54))

= -0.98

NOw we have to find p-value and critical valur for taking decision.

We can find p-value and critical value using EXCEL.

syntax :

P-value :

=NORMSDIST(z)

Where z is test statistic.

P-value = 0.1645

Critical value :

=NORMSINV(probability)

where probability = alpha/2

Critical value = -1.645

Here we see that P-value > alpha and test statistic < critical value therefore fail to reject H0 at 0.10 level of significance.

Conclusion : There is not suffiicient evidence to say that population mean of all such employees weights is less than 200 lb.

p q p^-p (p*q)/n sqrt test statistic 0.9 0.1 0.06 0.0009 0.03 2 0.928 0.072 0.032 0.000668 0.025848791 1.237968921 0.935 0.065 0.025 0.000608 0.024652586 1.014092393 0.94 0.06 0.02 0.000564 0.023748684 0.842151921 0.95 0.05 0.01 0.000475 0.021794495 0.458831468 0.96 0.04 0 0.000384 0.019595918 0 0.97 0.03 -0.01 0.000291 0.017058722 -0.58621038 0.98 0.02 -0.02 0.000196 0.014 -1.42857143 0.981 0.019 -0.021 0.000186 0.013652472 -1.53818294 0.982 0.018 -0.022 0.000177 0.013295112 -1.6547435 0.983 0.017 -0.023 0.000167 0.012927103 -1.77920756

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Solve the problem. 3) A manufacturer finds that in a random sample of 100 of its

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