Question 1197932: . The National Cable and Telecommunications Association recently reported that
the mean number of HDTVs per household in the United States is 2.30 with a standard
deviation of 1.474 sets. A sample of 100 homes in Boise, Idaho, revealed the following
sample information.
Tutorial #21
in Connect
Final PDF to printer
NONPARAMETRIC METHODS: NOMINAL LEVEL HYPOTHESIS TESTS 577
lin39470_ch15_542-578.indd 577 10/09/19 07:51 AM
Number of HDTVs Number of Households
0 7
1 27
2 28
3 18
4 10
5 or more 10
Total 100
At the .05 significance level, is it reasonable to conclude that the number of HDTVs per
household follows a normal distribution? (Hint: Use limits such as 0.5 up to 1.5, 1.5 up to
2.5, and so on.)
Answer by onyulee(41)   (Show Source):
You can put this solution on YOUR website! **1. Define Hypotheses**
* **Null Hypothesis (H0):** The number of HDTVs per household in Boise follows a normal distribution with a mean of 2.30 and a standard deviation of 1.474.
* **Alternative Hypothesis (H1):** The number of HDTVs per household in Boise does not follow a normal distribution.
**2. Determine Expected Frequencies**
* **Divide the data into intervals:**
* 0.5 - 1.5
* 1.5 - 2.5
* 2.5 - 3.5
* 3.5 - 4.5
* 4.5 - 5.5
* 5.5 and above
* **Calculate the z-scores for the interval boundaries:**
* For example, for the first interval (0.5 - 1.5):
* z1 = (0.5 - 2.30) / 1.474 = -1.22
* z2 = (1.5 - 2.30) / 1.474 = -0.54
* **Use the standard normal distribution table (z-table) to find the area under the curve for each interval.**
* **Multiply the area under the curve for each interval by the sample size (100) to get the expected frequency for that interval.**
**3. Calculate the Chi-Square Test Statistic**
* **For each interval:**
* Calculate the difference between the observed frequency and the expected frequency.
* Square the difference.
* Divide the squared difference by the expected frequency.
* **Sum the values calculated for each interval.** This sum is the chi-square test statistic.
**4. Determine the Degrees of Freedom**
* Degrees of freedom (df) = k - p - 1
* where k is the number of intervals (6 in this case)
* and p is the number of parameters estimated from the sample (0 in this case, as we are using the population mean and standard deviation)
* df = 6 - 0 - 1 = 5
**5. Find the Critical Value**
* Use a chi-square distribution table to find the critical value at the 0.05 significance level with 5 degrees of freedom.
**6. Compare the Test Statistic to the Critical Value**
* If the calculated chi-square test statistic is greater than the critical value, reject the null hypothesis.
* If the calculated chi-square test statistic is less than or equal to the critical value, fail to reject the null hypothesis.
**7. Conclusion**
* If you reject the null hypothesis, you can conclude that the number of HDTVs per household in Boise does not follow a normal distribution.
* If you fail to reject the null hypothesis, you cannot conclude that the number of HDTVs per household in Boise does not follow a normal distribution.
**Note:**
* This is a general outline of the process. You would need to use statistical software or a calculator to perform the calculations and find the critical value.
* This analysis assumes that the sample is representative of the population of households in Boise.
**Disclaimer:** This explanation provides a general framework. The specific calculations and interpretations may vary depending on the software used and the exact values obtained.
|
|
|
