Understanding Basic Statistics
Understanding Basic Statistics
8th Edition
ISBN: 9781337558075
Author: Charles Henry Brase, Corrinne Pellillo Brase
Publisher: Cengage Learning
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Textbook Question
Chapter 11.2, Problem 9P

For Problems 5-14, please provide the following information.

(a) What is the level of significance? State the null and alternate hypotheses.

(b) Find the value of the chi-square statistic for the sample. Are all the expected frequencies greater than 5? What sampling distribution will you use? What are the degrees of freedom?

(c) Find or estimate the P-value of the sample test statistic.

(d) Based on your answers in parts (a)-(c), will you reject or fail to reject the null hypothesis that the population fits the specified distribution of categories?

(e) Interpret your conclusion in the context of the application

Meteorology: Normal Distribution The following problem is based on information from the National Oceanic and Atmospheric Administration (NOAA) Environment Data Service. Let be a random variable that represents the average daily temperature (in degrees Fahrenheit ) in July in the town of Kit Carson, Colorado. The x distribution has a mean μ of approximately 75°F and standard deviation σ of approximately 8°F . A 20-year study (620 July days) gave the entries in the rightmost column of the following table.

I II III IV
Region under Normal Curve x ° F Expected % from Noraml Curve Observed Number of Days in 20 Years
µ 3 σ x < µ 2 σ 51 x < 59 2.35% 16
µ 2 σ x < µ σ 59 x < 67 13.5% 78
µ σ x < µ 67 x < 75 34% 212
µ x < µ + σ 75 x < 83 34% 221
µ + σ x < µ + 2 σ 83 x < 91 13.5% 81
µ + 2 σ x < µ + 3 σ 91 x < 99 2.35% 12

(i) Remember that μ  = 75 and  σ  = 8 . Examine Figure 7-3 in Chapter 7. Write a brief explanation fur Columns I, II, and III in the context of this problem.

(ii) Use a 19 level of significance to test the claim that the average daily. July temperature follows a normal distribution with μ  = 75 and   σ = K .

(i)

Expert Solution
Check Mark
To determine

To explain: The columns I, II and III in the context of this problem.

Explanation of Solution

Let, X represents the average daily temperature (in degrees Fahrenheit), which follows the normal distribution with μ=75 and σ=8.

The below graph shows normal curve with μ=75 and σ=8.

We can see that, 2.35% of the data values will lie within 2 standard deviation below the mean and 3 standard deviation below the mean, 13.55% of the data values will lie within 1 standard deviation below the mean and 2 standard deviation below the mean, 34.0% of the data values will lie within mean and 1 standard deviation below the mean and mean.

Understanding Basic Statistics, Chapter 11.2, Problem 9P

By empirical rule, approximately 68% of the data values lie within 1 standard deviation on each side of the mean, approximately 95% of the data values lies within 2 standard deviations on each side of the mean and approximately 99.7% of the data values lies within 3 standard deviations on each side of the mean.

(ii)

(a)

Expert Solution
Check Mark
To determine

The level of significance and state the null and alternative hypotheses.

Answer to Problem 9P

Solution: The level of significance is 0.01. H0: The average daily July temperature follows a normal distributionand H1: The average daily July temperature does not follow a normal distribution.

Explanation of Solution

The level of significance, α=0.01.

The null hypothesis for testing is defined as,

H0: The average daily July temperature follows a normal distribution.

The alternative hypothesis is defined as,

H1: The average daily July temperature does not follow a normal distribution.

(b)

Expert Solution
Check Mark
To determine

To test: The value of chi-square statistic for the sample, whether all the expected frequencies are greater than 5 and also explain the sampling distribution to be used and find degrees of freedom.

Answer to Problem 9P

Solution: The value of chi-square statistic for the sample is 1.5590. Expected frequencies are greater than 5. We have to use chi-square distribution and degrees of freedom are 5.

Explanation of Solution

Calculation: To find the χ2 statistic for the sample by using MINITAB software is as:

Step 1: Go to Stat > Tables > Chi-square Goodness-of-fit-test.

Step 2: Select ‘Days_20_yrs’ in ‘Observed counts’.

Step 3: Choose ‘Normal_curve’ in ‘Proportions specified by Historical counts’. Then click on OK.

The Minitab output is:

Chi-Square Goodness-of-Fit Test

Category Observed Historical
Counts
Test
Proportion
Expected Contribution
to Chi-Square
1 16 0.0235 0.023571 14.614 0.131481
2 78 0.1350 0.135406 83.952 0.421963
3 212 0.3400 0.341023 211.434 0.001514
4 221 0.3400 0.341023 211.434 0.432771
5 81 0.1350 0.135406 83.952 0.103791
6 12 0.0235 0.023571 14.614 0.467513

Chi-Square Test

N DF Chi-Sq P-Value
620 5 1.55903 0.906

Therefore, the obtained Chi-square test statistics is 1.5590.

The obtained expected frequencies are:

Expected
Frequencies
14.614
83.952
211.434
211.434
83.952
14.614

Therefore, all expected frequencies are greater than 5.

The chi-square distribution will use in this study and the degrees of freedom is 5.

Conclusion: χ2statistic1.559 and expected frequencies are greater than 5. We have to use chi-square distribution and degrees of freedom are 5.

(c)

Expert Solution
Check Mark
To determine

The P-value of the sample statistic.

Answer to Problem 9P

Solution: The P-value of sample statistic is 0.906.

Explanation of Solution

Calculation: The Minitab output obtained in above part (b) also gives the P-value. So, the estimate P-valuefor the sample test statistic is 0.906.

(d)

Expert Solution
Check Mark
To determine

Whether we reject or fail to reject the null hypothesis.

Answer to Problem 9P

Solution: We failed to reject the null hypothesis is at 1% level of significance.

Explanation of Solution

The obtained results in part (a), (b) and (c) are,

α=0.01

χ2statistic1.5590 with5 degrees of freedom.

Since, the P-value(0.906) is less than 0.01, hence, we failed to reject the null hypothesis at α=0.01. Therefore we can conclude that the data are not statistically significant at 1% level of significance.

(e)

Expert Solution
Check Mark
To determine

To explain: The conclusion in the context of application.

Answer to Problem 9P

Solution: We have insufficient evidence to conclude that the average daily July temperature does not follow a normal distribution at 1% level of significance.

Explanation of Solution

From above part, we failed to reject the null hypothesis of independence at α=0.01. Therefore, we can conclude that the data are not statistically significant at 1% level of significance.

We have insufficient evidence to conclude that the average daily July temperature does not follow a normal distributionat 1% level of significance.

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Chapter 11 Solutions

Understanding Basic Statistics

Ch. 11.1 - For Problems 9-19, please provide the following...Ch. 11.1 - For Problems 9-19. please provide the following...Ch. 11.1 - For Problems 9-19. please provide the following...Ch. 11.1 - For Problems 9-19. please provide the following...Ch. 11.1 - For Problems 9-19. please provide the following...Ch. 11.1 - For Problems 9-19. please provide the following...Ch. 11.1 - Prob. 17PCh. 11.1 - Prob. 18PCh. 11.1 - Prob. 19PCh. 11.2 - Statistical Literacy For a chi-square...Ch. 11.2 - Statistical Literacy How are expected frequencies...Ch. 11.2 - Statistical Literacy Explain why goodness-of-fit...Ch. 11.2 - Critical Thinking When the sample evidence is...Ch. 11.2 - For Problems 5-14, please provide the following...Ch. 11.2 - For Problems 5-14, please provide the following...Ch. 11.2 - For Problems 5-14, please provide the following...Ch. 11.2 - For Problems 5-14, please provide the following...Ch. 11.2 - For Problems 5-14, please provide the following...Ch. 11.2 - Prob. 10PCh. 11.2 - Prob. 11PCh. 11.2 - For Problems 5-14, please provide the following...Ch. 11.2 - Prob. 13PCh. 11.2 - Prob. 14PCh. 11.2 - Prob. 15PCh. 11.2 - For Problems 5-14, please provide the following...Ch. 11.3 - Statistical Literacy Docs the x distribution need...Ch. 11.3 - Prob. 2PCh. 11.3 - Prob. 3PCh. 11.3 - For Problems 3-11, please provide the following...Ch. 11.3 - For Problems 3-11. please provide the following...Ch. 11.3 - For Problems 3-11. please provide the following...Ch. 11.3 - Prob. 7PCh. 11.3 - For Problems 3-11, please provide the following...Ch. 11.3 - For Problems 3-11. please provide the following...Ch. 11.3 - Prob. 10PCh. 11.3 - Prob. 11PCh. 11.4 - Prob. 1PCh. 11.4 - Statistical Literacy What is the symbol used for...Ch. 11.4 - Prob. 3PCh. 11.4 - Statistical Literacy How does the t value for the...Ch. 11.4 - Prob. 5PCh. 11.4 - Using Computer Printouts Problems 5 and 6 use the...Ch. 11.4 - In Problems 7-12, parts (a) and (b) relate to...Ch. 11.4 - In Problems 7-12, parts (a) and (b) relate to...Ch. 11.4 - In Problems 7-12, parts (a) and (b) relate to...Ch. 11.4 - In Problems 7-12, parts (a) and (b) relate to...Ch. 11.4 - Prob. 11PCh. 11.4 - Prob. 12PCh. 11.4 - Prob. 13PCh. 11.4 - Prob. 14PCh. 11.4 - Prob. 15PCh. 11.4 - Prob. 16PCh. 11.4 - Prob. 17PCh. 11.4 - Prob. 18PCh. 11.4 - Prob. 19PCh. 11 - Terminology Match each of the following tests to...Ch. 11 - Prob. 2CRCh. 11 - Statistical Literacy of the following random...Ch. 11 - Prob. 4CRCh. 11 - Prob. 5CRCh. 11 - Before you solve Problems 6-10, first classify the...Ch. 11 - Prob. 7CRCh. 11 - Before you solve Problems 6-10, first classify the...Ch. 11 - Before you solve Problems 6-10, first classify the...Ch. 11 - Prob. 10CRCh. 11 - Prob. 11CRCh. 11 - Prob. 12CRCh. 11 - Prob. 13CRCh. 11 - Prob. 14CRCh. 11 - Prob. 15CRCh. 11 - The Statistical Abstract of the United States...Ch. 11 - Prob. 1LCWPCh. 11 - Prob. 2LCWPCh. 11 - Prob. 3LCWPCh. 11 - Prob. 4LCWPCh. 11 - Prob. 1CRPCh. 11 - Prob. 2CRPCh. 11 - Prob. 3CRPCh. 11 - Prob. 4CRPCh. 11 - Prob. 5CRPCh. 11 - Prob. 6CRPCh. 11 - Prob. 7CRP
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