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Could you check my answer on this <span title="statistics/probability/hypothesis">statistics/probability/hy...</span> question?

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Patients recovering from an appendix operation normally spend an average of 6.3 days in the hospital. The distribution of recovery times is normal with a s.d. of 2 days. The hospital is trying a new recovery program designed to lessen the time patients spend in the hospital. The first 16 appendix patients in this new program were released from the hospital in an average of 5.3 days. On the basis of these data, can the hospital conclude that the new program has a significant reduction of recovery time? Test at the .05 level of significance.

The appropriate statistical procedure for this example would be a

a. t-test

b. z-test

(I chose B)

Is this a one-tailed or a two-tailed test?

(I said one-tailed)

The most approximate null hypothesis would be:

A. μ new program = 6.3

B. μ new program = 5.5

C. μ new program ≤ 6.3

D. μ new program ≥ 6.3

(I said C)

Setup the criteria for making a decision. That is, find the critical value using an alpha = 0.5 using your tables.

(I said -0.5.)

What is the numberic value of your standard error?

0.5

What is the t- or z-value you obtained?

.4801

Based on your results, you would

A. reject the null hypothesis

B. fail to reject the null hypothesis

(I said A)

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1 ANSWERS


  1. The most approximate null hypothesis would be:

    A. μ new program = 6.3

    *this is because you&#039;re supposed to reject the null if something has changed. alternative hypothesis would be μ ≤ 6.3

    Setup the criteria for making a decision. That is, find the critical value using an alpha = 0.5 using your tables.

    Critical value = -z(0.05) = -1.645

    *It&#039;s a lower tail rejection region

    What is the numberic value of your standard error?

    standard error = 2/(16)^0.5 = 2/4 = 0.5

    What is the t- or z-value you obtained?

    z-value = (5.3 - 6.3)/0.5 = -2

    Based on your results, you would

    -2&lt;-1.645. Thus (A)

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