Examples for Hypothesis Testing

  1. A newspaper claims that over 40% of college undergrads have wild parties before they obtain their bachelor’s degree
  1. State Null hypothesis, State Alternative Hypothesis
  2. A statistician performs study and the proportion from the sample he was working with turned out to be 0.45. The probability of observing a value of 0.45 or more (p-value) in such sample is 0.20 when assuming that null hypothesis is true. Do you have strong evidence to reject the Null Hypothesis?

a) Ho: p=.40

Ha: p > 0.40

2) p0 = 0.40

phat = 0.45

p-vaue = 0.20

Since the level of significance is not mentioned it is assumed to be 0.05, so we would reject H0 if p-value < 0.05. which is not the case here. Conclusion: we FAIL TO REJECT null hypothesis concluding that there is no enough evidence to say that the proportion of undergrads is more than 0.40

  1. It is important for travelers that mass transit vehicles exhibit a high percentage of on-time performance. Suppose a regional train system boasts that its trains run on time over 80% of the time. To check out this claim investigator collected the sample of real data, and sample of 40 trains arrivals was observed 37 of which came on time.
  1. What’s population proportion in this problem
  2. What claim is being made about this population proportion?
  3. Stat null and alternative hypothesis
  4. Would you reject or not Ho?

Ho: p=.80

Ha: p > 0.80

phat=37/40 = .925

po = .80

z-statistic = (phat - po) / sqrt(po (1-po)/n) = (.925 - .80)/sqrt(.80(.20)/40) = .125/.0632 = 1.97

Since testing Ha: p >.80, p-value is the area in the upper tail, or to the right of z-value.

Looking at the Table, find percentile associated with z=1.97 = 2.0  97.73%, but that’s area to the left, since we need area to the right we do: 1- .9773 = 0.0227

p-value = 0.0227

By default we compare p-value with 0.05 significance level and see whether

p-value < 0.05? Yes it is. Conclusion: We reject Ho, stating that there is enough evidence to say that trains run on time more than 80% of time.

  1. Records from several year ago show that in a certain community 60% of the people owned their houses. An advocacy group for affordable housing claims that home ownership has declined since that time. A survey of 50 families show that 9 own their home. Is this conclusive evidence that there has been a decrease in home ownership in the community?

Ho: p=0.60

Ho: p < 0.60

phat = 9/50 = 0.18

po= 0.60

z-value = (phat - po) / sqrt(po (1-po)/n) = (0.18 – 0.60) / (sqrt(0.60*0.40/50)) =

- 0.42/ 0.0692 = - 6.069

Perventile associated with z= - 6 is not in the Table B. For all the values of Z< - 3.4 the percentiles will be less than 0.03%, . We will take 0.02% for z = - 6

Since our Alternative hypothesis has sign ‘<’ meaning that p-value is in the left (lower) tail, the p-value is 0.02% / 100 =0.0002 which is very small compared to 0.05 significance level. Conclusion: we REJECT Null hypothesis giving evidence that there has been a decrease in home ownership.

  1. Public Health Service examined a representative cross section of several thousand men aged 18 to 74. The systolic blood pressure of these men followed normal distribution with mean mu =129 and standard deviation, sigma = 18. Suppose we took a random sample of 44 women aged 18 to 74 and found that their average systolic blood pressure is xbar = 120 with st dev s = 21. Does this prove that, on average women have lower systolic blood pressure than men?

Ha: mu <129

H0: mu = 129

mu0 = 129

xbar = 120

s=21

n =44

z-value = (xbar – mu0)/(s/sqrt(n)) = (120 – 129) / 21/sqrt(44) = - 2.84

p-value = 0.26%/100 = 0.0026 since it is in lower tail.

p-value is less than 0.05, meaning that we REJECT hull hypothesis, concluding that there is strong evidence to say that on average women have lower systolic blood pressure.