Hypothesis test involving m, s unknown

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Hypothesis test involving m,s unknown

• We have seen how to conduct hypothesis tests
  involving µ when we have a normal distribution
  and known s, using the z distribution.
• We now turn to the situation where s is unknown
  but has a normal distribution (because the
  sample size is large or the sample population is
  normal).
• Since s is unknown, we use s in its place.
• However, the random variable does not
  have a have a standard normal distribution so we
  do not call it z and we cannot refer to the z
  table for comparison.
• Instead, the we will use the t distribution, and
  the test statistic will be

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• Example A random sample of 25 students
  registering for classes showed the mean waiting
  time in the registration line was 22.6 minutes
  and the standard deviation was 8.0 minutes. Is
  there any evidence to support the student
  newspapers claim that registration time takes
  longer than 20 minutes? Use a 0.05 and assume
  waiting time is approximately normal.
• Solution
• 1. State the null and alternative hypotheses
• H0 m 20 () (no longer than)
• Ha m gt 20 (longer than)

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• Determine the type of test The sampled
  population is approximately normal. s is
  unknown. Use t with df n - 1 24
• Define the Rejection RegionThis is a
  right-tailed test with a.05, so we reject H0 if
  .
• Calculate the test statistic
• State the conclusion We do not reject H0, so we
  do not have sufficient evidence to support the
  student newspapers claim that registration time
  takes longer than 20 minutes, in other words, we
  will assume the true registration time is less
  than or equal to 20 minutes.

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• Using the p-value procedure
• Note
• 1. If this hypothesis test is done with the aid
  of a computer a fairly precise p-value can be
  calculated. Using the table, we are somewhat
  limited by the probabilities available.
• We can use the t-table to place bounds on the
  p-value.0.05 lt p lt 0.10
• We can use software to calculate a p-value.The
  p-value is not smaller than the level of
  significance, a.

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• Hypothesis-Testing Procedure for population
  proportion
• For hypothesis tests concerning the binomial
  parameter p, use the test statistic z
• Example (Probability-Value Approach) A hospital
  administrator believes that at least 75 of all
  adults have a routine physical once every two
  years. A random sample of 250 adults showed 172
  had physicals within the last two years. Is
  there any evidence to refute the administrator's
  claim? Use a 0.05.

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• Solution
• 1. State the null and alternative hypotheses
• H0 p 0.75
• Ha p lt 0.75
• 2. Determine the type of test
• Underlying binomial distribution.
• Test statistic z.
• n 250gt30
• np (250)(0.75) 187.5 gt 5
• nq (250)(0.25) 62.5 gt 5
• Level of significance a 0.05

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• 3. Rejection Region Reject H0 if plt.05.
• 4. Calculating the test statistic
• a. Sample information
• b. The test statistic

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• b. The p-value is smaller than the level of
  significance, a.
• 5. Conclusion There is evidence to suggest the
  proportion of adults who have a routine physical
  exam every two years is less than 0.75.

p-value
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• Example A university bookstore employee in
  charge of ordering texts believes 65 of all
  students sell their statistics book back to the
  bookstore at the end of the class. To test this
  claim, 200 statistics students are selected at
  random and 141 plan to sell their texts back to
  the bookstore. Is there any evidence to suggest
  the proportion is different from 0.65? Use a
  0.01.
• Solution
• 1. The null and alternative hypotheses
• H0 p 0.65
• Ha

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• 2. Determine the type of test
• Underlying binomial distribuiton
• Test statistic z
• n 200
• np (200)(0.65) 130 gt 5 nq
  (200)(0.35) 70 gt 5
• Level of significance a 0.01
• 3. Rejection Region Reject H0 if zgt2.58 or
  zlt-2.85
• 4. Calculate the value of the test statistic

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• 5. Conclusion There is no evidence to suggest
  the true proportion of students who sell their
  statistics text back to the bookstore is
  different from 0.65.

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• Note
• 1. There is a relationship between confidence
  intervals and two-tailed hypothesis tests when
  the level of confidence and the level of
  significance are complementary.
• 2. The confidence interval and the width of the
  noncritical region are the same.
• 3. The point estimate is the center of the
  confidence interval, and the hypothesized mean is
  the center of the noncritical region.
• 4. If the hypothesized value of p is contained in
  the confidence interval, then the test statistic
  will be in the noncritical region.
• 5. If the hypothesized value of p does not fall
  within the confidence interval, then the test
  statistic will be in the critical region.