Question

1. A custodian wishes to compare two competing floor waxes to decide which one is best. He believes that the mean of WaxWin is not equal to the mean of WaxCo. In a random sample of 37 floors of WaxWin and 30 of WaxCo. WaxWin had a mean lifetime of 26.2 and WaxCo had a mean lifetime of 21.9. The population standard deviation for WaxWin is assumed to be 9.1 and the population standard deviation for WaxCo is assumed to be 9.2. Perform a hypothesis test using a significance level of 0.10 to help him decide. Let WaxWin be sample 1 and WaxCo be sample 2. The correct hypotheses are: H 0 : μ 1 ≤ μ 2 H 0 : μ 1 ≤ μ 2 H A : μ 1 > μ 2 H A : μ 1 > μ 2 (claim) H 0 : μ 1 ≥ μ 2 H 0 : μ 1 ≥ μ 2 H A : μ 1 < μ 2 H A : μ 1 < μ 2 (claim) H 0 : μ 1 = μ 2 H 0 : μ 1 = μ 2 H A : μ 1 ≠ μ 2 H A : μ 1 ≠ μ 2 (claim) Correct

Since the level of significance is 0.10 the critical value is 1.645 and -1.645

The test statistic is: Incorrect(round to 3 places)

The p-value is: Incorrect(round to 3 places)

A random sample of 30 chemists from Washington state shows an average salary of $42546, the population standard deviation for chemist salaries in Washington state is $868. A random sample of 39 chemists from Florida state shows an average salary of $48395, the population standard deviation for chemist salaries in Florida state is $945. A chemist that has worked in both states believes that chemists in Washington make more than chemists in Florida. At αα=0.05 is this chemist correct?

Let Washington be sample 1 and Florida be sample 2.

The correct hypotheses are:

• H0:μ1≤μ2H0:μ1≤μ2
  HA:μ1>μ2HA:μ1>μ2(claim)
• H0:μ1≥μ2H0:μ1≥μ2
  HA:μ1<μ2HA:μ1<μ2(claim)
• H0:μ1=μ2H0:μ1=μ2
  HA:μ1≠μ2HA:μ1≠μ2(claim)

Since the level of significance is 0.05 the critical value is 1.645
The test statistic is: (round to 3 places)
The p-value is: (round to 3 places)

A researcher is interested in seeing if the average income of rural families is greater than that of urban families. To see if his claim is correct he randomly selects 45 families from a rural area and finds that they have an average income of $66299 with a population standard deviation of $668. He then selects 31 families from a urban area and finds that they have an average income of $67979 with a population standard deviation of $534. Perform a hypothesis test using a significance level of 0.01 to test his claim. Let rural families be sample 1 and urban familis be sample 2.

The correct hypotheses are:

• H0:μ1≤μ2H0:μ1≤μ2
  HA:μ1>μ2HA:μ1>μ2(claim)
• H0:μ1≥μ2H0:μ1≥μ2
  HA:μ1<μ2HA:μ1<μ2(claim)
• H0:μ1=μ2H0:μ1=μ2
  HA:μ1≠μ2HA:μ1≠μ2(claim)

Since the level of significance is 0.01 the critical value is 2.326
The test statistic is: (round to 3 places)
The p-value is: (round to 3 places)

A researcher is interested in seeing if the average income of rural families is greater than that of urban families. To see if his claim is correct he randomly selects 45 families from a rural area and finds that they have an average income of $66299 with a population standard deviation of $668. He then selects 31 families from a urban area and finds that they have an average income of $67979 with a population standard deviation of $534. Perform a hypothesis test using a significance level of 0.01 to test his claim. Let rural families be sample 1 and urban familis be sample 2.

The correct hypotheses are:

• H0:μ1≤μ2H0:μ1≤μ2
  HA:μ1>μ2HA:μ1>μ2(claim)
• H0:μ1≥μ2H0:μ1≥μ2
  HA:μ1<μ2HA:μ1<μ2(claim)
• H0:μ1=μ2H0:μ1=μ2
  HA:μ1≠μ2HA:μ1≠μ2(claim)

Since the level of significance is 0.01 the critical value is 2.326
The test statistic is: (round to 3 places)
The p-value is: (round to 3 places)

Answer 1

1) H₀:

    H_A:

The test statistic z = ()/

                            = (26.2 - 21.9)/sqrt((9.1)^2/37 + (9.2)^2/30)

                            = 1.912

P-value = 2 * P(Z > 1.912)

           = 2 * (1 - P(Z < 1.912))

          = 2 * (1 - 0.972)

        = 0.056

2) H₀:

    H_A:

The test statistic z = ()/

                            = (42546 - 48395)/sqrt((868)^2/30 + (945)^2/39)

                            = -26.694

P-value = P(Z > -26.694)

            = 1 - P(Z < -26.694)

            = 1 - 0 = 1

3) H₀:

    H_A:

The test statistic z = ()/

                            = (66299 - 67979)/sqrt((668)^2/45 + (534)^2/31)

                            = -12.151

P-value = P(Z > -12.151)

         = 1 - P(Z < -12.151)

        = 1 - 0 = 1