Probability should be rounded to 4 decimal places. Mean and standard deviations

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Probability should be rounded to 4 decimal places. Mean and standard deviations to 2 decimal places.

Chapter 5 (Normal Distributions # 1-3)

2.) Proctor and Gamble reported that an American family of four washes an average of 2560 pounds of clothes each year. If the standard deviation of the distribution is 261.5 pounds, find the probability that the mean of a randomly selected sample of 65 families of four will be between 2617 and 2799 pounds. (Hint: Use the Central Limit Theorem)

3.) X is a normally distributed random variable with a mean of 11.7. Find the standard deviation of the distribution if 59.10% of the data lies to the right of 10.09. Chapter 6 (Confidence Intervals # 6-8)

4.) A medical researcher wishes to determine the percentage of females who take vitamins. He wishes to be 95% confident that the estimate is within 2.35 percentage points of the true proportion. A recent study of 315 females showed that 68% took vitamins. How large should the sample size be? (Page 324)

5.) A random sample of 200 four year olds attending day care centers provided a yearly tuition average of $6237 and a population standard deviation of $315. Find the 99% confidence interval of the true mean. To earn FULL credit, include an interpretation statement with your answer. (Section 6.1)

6.) The following data are the calories per half-cup serving for 16 popular chocolate ice cream brands. 300 350 275 180 260 395 185 195 260 170 160 290 170 180 225 145 Estimate the true mean calories per half-cup serving of chocolate ice cream with 90% confidence. To Earn FULL credit, Include an interpretation statement with your answer. (Section 6.2) Chapter 7 (Hypothesis Test #7 - 8) – Show all steps – See page 357 for list of steps - for each hypothesis test in order to earn FULL credit.

7.) An educator claims that the average salary of substitute teachers in school districts in TN earn more than $70 per day. A random sample of eight districts is selected, and the daily salaries are given below. With = 0.05, is there enough evidence to support the claim? 72 65 50 66 80 76 65 68

8.) A telephone company representative estimates that 90% of its customers have call waiting service. To test this hypothesis, she selected a sample of 150 customers and found that 86% had call waiting. With = 0.01, is there enough evidence to reject the claim? Chapter 8 ( Hypothesis Test # 9)

9.) A survey found that the average hotel room rate in New Orleans is $191.67 and the average room rate in Phoenix is $189.67. Assume that the data were obtained from two samples of 74 hotels each and that the standard deviation of the populations is $22.30 and $21.45, respectively. At 0.05, can it be concluded that there is a significant difference in the rates?

Explanation / Answer

Let X be the random variable that number of four washes.

X ~ N(mean = 2560 pounds, sd = 261.5 pounds)

Find the probability that the mean of a randomly selected sample of 65 families of four will be between 2617 and 2799 pounds.

n = sample size = 65

And we have to find P(2617 < Xbar < 2799).

First convert xbar = 2617 and xbar =2799 into z-score.

But we know that the distribution of sample mean is,

Xbar ~ N(mean = 2560 pounds, sd = 261.5 / sqrt(65))

Xbar ~ N(mean = 2560 pounds, sd = 32.44)

z-score for xbar = 2617 is,

x = (2617 - 2560) / 32.44 = 1.76

z-score for x= 2799 is,

z = (2799 - 2560) / 32.44 = 7.37

That is now we have to find P(1.76 < Z < 7.37).

P(1.76 < Z < 7.37) = P(Z <=7.37) - P(Z<=1.76)

These probabilities we can find by using EXCEL.

syntax is,

=NORMSDIST(z)

where z is z-score value.

P(Z <=7.37) = 1

P(Z<=1.76) = 0.9606

P(1.76 < Z < 7.37) = 1 - 0.9606 = 0.0394

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n = 200

Xbar = $6237

= $315

And we have to find 99% confidence interval for true mean.

Here we use Z-interval because population standard deviation is given.

C-level = 99% = 0.99

We can find the confidence interval by using TI-83 calculator.

steps:

STAT --> TESTS --> ZInterval --> ENTER --> Highlight on Stats --> ENTER --> Input value of , Xbar,n and C-level --> Calculate --> ENTER

Output is,

99% confidence interval for true mean is (6179.6, 6294.4).

We are 99% confident that the true population mean is lies between 6179.6 and 6294.4.

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Here we have to test the hypothesis that,

H0 : mu = $70 Vs H1 : mu > $70

where mu is population mean.

Here we use one sample t-test because sample size is small and population standard deviation is unknown.

n = sample size = 8

alpha = level of significance = 0.05

This also we can done by using TI_83 calculator.

steps :

STAT --> ENTER --> Enter all the data in L1 --> STAT --> TESTS --> 2: T-Test --> ENTER --> Highlight on Data --> ENTER --> Mu0 = 70, List : L1, Freq:1 --> Select alternative ">" --> Calculate --> ENTER

t = -0.7056

P-value = 0.748

P-value > alpha

Accept H0 at 5% level of significance.

Conclusion : There is not sufficient evidence to say that average salary of substitute teachers in school districts in TN earn more than $70 per day.

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Here we have to test the hypothesis that,

H0 : p = 90% = 0.9 Vs H1 : p 90% = 0.9

where p is population proportion.

n = 150

p^ = 86% = 0.86

alpha = level of significance = 1% = 0.01

The test statistic is,

Z = (p^ - p) / sqrt(pq / n)

Z = (0.86 - 0.9) / sqrt(0.9*0.1 / 150)

Z = -0.04 / sqrt(0.0006)

Z = -0.04 / 0.0245

Z = -1.63

For taking decision we have to find P-value.

EXCEL syntax for P-value is,

=NORMSDIST(z)

where z is test statistic value.

P-value = 0.051

P-value > alpha

Accept H0 at 1% level of significance.

Conclusion : There is sufficient evidence to say that A telephone company representative estimates that 90% of its customers have call waiting service.

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Here we have to test the hypothesis that,

H0 : mu1 = mu2 vs H1 : mu1 mu2

where mu1 is population mean for room rate in New Orleans and

mu2 is population mean for room rate in Phoenix.

alpha = 5% = 0.05

X1bar = $191.67

X2bar = $189.67

1 = $22.30

2 = $21.45

n1 = 74

n2 = 74

TI-83 steps for two sample Z-test are,

STAT --> TESTS --> 3:2-SampZTest --> ENTER --> Highlight on Stats --> ENTER --> Input all the values --> Select alternative "" --> Calculate --> ENTER

Output is :

Z = 0.5560

P-value = 0.578

P-value > alpha

Accept H0 at 5% level of significance.

Conclusion : There is sufficient evidence to say that two population means are equal.

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Probability should be rounded to 4 decimal places. Mean and standard deviations

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Probability should be rounded to 4 decimal places. Mean and standard deviations

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