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    Hypothesis Testing

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    FIRST

    In 1990, 5.8% of job applicants who were tested for drugs failed the test. At the 0.01 level, test the claim that the failure rate is not lower if a random sample of 1520 current job applicants results in 58 failures. Work and steps must be shown.

    Write each step of hypothesis procedure

    SECOND
    The expense of moving the storage yard for the Consolidated Package Delivery Service is justified only if it can be shown that the daily mean travel distance will be less than 214 mi. In trial runs of 12 delivery trucks, the mean and standard deviation are found to be 198 mi and 42, respectively. At the 0.01 level of significance, test the claim that the mean is less than 214 mi. Assume that the data are normally distributed. Write each step of the hypothesis procedure. Round test statistic to hundredths. Work and steps must be shown.

    Write each step of hypothesis procedure

    THIRD
    Nielson Media Research reported that adult women watch TV an average of 5 hours and 1 minute per day, compared to an average of 4 hours and 17 minutes for adult men. Assume that those results are found from a sample of 100 men and 100 women and that the two groups have the same standard deviation of 57 minutes. At the 0.05 level of significance, test the claim that there is not a gender difference in the time spent watching television. Assume that the data are normally distributed. Write each step of the hypothesis procedure. Round test statistic to hundredths. Work and steps must be shown.

    Write each step of hypothesis procedure

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    FIRST
    <br>
    <br>In 1990, 5.8% of job applicants who were tested for drugs failed the test. At the 0.01 level, test the claim that the failure rate is not lower if a random sample of 1520 current job applicants results in 58 failures. Work and steps must be shown.
    <br>Write each step of hypothesis procedure
    <br>
    <br>The claimed proportion is p=0.058
    <br>The sample proportion is x=58/1520=0.03816
    <br>Ho: p=0.058
    <br>Ha: p<0.058
    <br>The standard error is SE=SQRT(p*(1-p)/n)=SQRT(0.058*0.942/1520)=0.006
    <br>Since x>p, we use the formula for z test:
    <br>Z = (x-p+0.5/n)/SE = (0.03816-0.058+0.5/1520)/0.006=-3.252
    <br>From the z-table, we find that 0.01 z critical value is 2.33
    <br>Since the absolute value of the calculated z (3.252) is greater than critical value 2.33, we have to reject Ho at the 0.01 level. Therefore, we conclude that the failure rate is lower than ...

    Solution Summary

    FIRST

    In 1990, 5.8% of job applicants who were tested for drugs failed the test. At the 0.01 level, test the claim that the failure rate is not lower if a random sample of 1520 current job applicants results in 58 failures. Work and steps must be shown.

    Write each step of hypothesis procedure

    SECOND
    The expense of moving the storage yard for the Consolidated Package Delivery Service is justified only if it can be shown that the daily mean travel distance will be less than 214 mi. In trial runs of 12 delivery trucks, the mean and standard deviation are found to be 198 mi and 42, respectively. At the 0.01 level of significance, test the claim that the mean is less than 214 mi. Assume that the data are normally distributed. Write each step of the hypothesis procedure. Round test statistic to hundredths. Work and steps must be shown.

    Write each step of hypothesis procedure

    THIRD
    Nielson Media Research reported that adult women watch TV an average of 5 hours and 1 minute per day, compared to an average of 4 hours and 17 minutes for adult men. Assume that those results are found from a sample of 100 men and 100 women and that the two groups have the same standard deviation of 57 minutes. At the 0.05 level of significance, test the claim that there is not a gender difference in the time spent watching television. Assume that the data are normally distributed. Write each step of the hypothesis procedure. Round test statistic to hundredths. Work and steps must be shown.

    Write each step of hypothesis procedure

    $2.19

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