Question

A random survey of 75 death row inmates revealed that the mean length of time on death row is 17.3 years with a standard deviation of 6.5 years. Conduct a hypothesis test to determine if the population mean time on death row could likely be 15 years.

• Part (a)

  Is this a test of one mean or proportion?

  test of one meantest of proportions    

• Part (b)

  State the null and alternative hypotheses.

  H₀: μ ≥ 15
  H_a: μ < 15H₀: μ ≤ 15
  H_a: μ > 15    H₀: μ ≠ 15
  H_a: μ = 15H₀: μ = 15
  H_a: μ ≠ 15H₀: μ = 15
  H_a: μ > 15

• Part (c)

  Is this a right-tailed, left-tailed, or two-tailed test?

  right-tailed testleft-tailed test    two-tailed test

• Part (d)

  What symbol represents the random variable for this test?

  nα    Xσμ

• Part (e)

  In words, define the random variable for this test.

  the proportion of time spent on death row for 75 inmatesthe amount of time spent on death row for 75 inmates    the mean time spent on death row for 75 inmatesthe standard deviation of time spent on death row for 75 inmatesthe number of inmates on death row

• Part (f)

  Is the population standard deviation known?

  YesNo    

  
  If so, what is it? (Enter your answer to one decimal place. If the standard deviation is not known enter DNE.)

• Part (g)

  Calculate the following.

  (i) Enter your answer to one decimal place.
  x =
  
  (ii) Enter your answer to one decimal place.
  s =
  
  (iii) Enter your answer as a whole number.
  n =

• Part (h)

  Which test should be used?

  Normal testStudent's t-test    

• Part (i)

  State the distribution to use for the hypothesis test. (Enter your answer in the form z or t_df where df is the degrees of freedom.)

• Part (j)

  Find the p-value. (Round your answer to four decimal places.)

• Part (k)

  At a pre-conceived

  α = 0.05,

  what do you determine for each of the following?(i) Decision:

  reject the null hypothesisdo not reject the null hypothesis    

  
  (ii) Reason for the decision:

  Since α > p-value, we reject the null hypothesis.Since α < p-value, we reject the null hypothesis.    Since α > p-value, we do not reject the null hypothesis.Since α < p-value, we do not reject the null hypothesis.

  
  (iii) Conclusion (write out in a complete sentence):

  There is sufficient evidence to conclude that the mean length of time on death row is not 15 years.There is not sufficient evidence to conclude that the mean length of time on death row is not 15 years.    

Answer 1

Given :

Sample size = n = 75

Sample mean = = 17.3

Standard deviation = s = 6.5

a) This a test of one mean test of one meantest of proportions .

b) Hypothesis test :

The null and alternative hypothesis is

Ho : μ = 15

H_a: μ ≠ 15

c) This is a two-tailed test .

d) represents the random variable for this test.

e) The mean time spent on death row for 75 inmatesthe standard deviation of time spent on death row for 75 inmatesthe number of inmates on death row.

f) No, the population standard deviation is not know.

DNE

g) Sample size = n = 75

Sample mean = = 17.3

Standard deviation = s = 6.5

h) Students t - test should be use.

I) the t - distribution to use for the hypothesis test.

Degree of freedom = df= n-1 = 75-1 = 74

t74

j) Test statistic :

t = - /s/√n

= 17.3-15/6.5/√75

= 3.064

P-value : P-value corresponding to t = 3.064 with df = 74 is

P-value = 2 * P(t > 3.064) = 0.0030

P-value = 0.0030

i) Reject the null hypothesis.

ii) Since > P-value , we Reject the null hypothesis.

iii) There is sufficient evidence to conclude that mean length of time on death row is not 15 years.