STAT 1000 Lecture Notes - Lecture 35: Null Hypothesis, Confidence Interval, Statistical Hypothesis Testing
10 Sep 2018
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Lecture 35
Example - Fast Food:
A fast food restaurant claims that the average waiting time in their drive through is less
than a minute. We record the drive thru waiting times for a random sample of 31
customers. The sample average time is 54.9 seconds and the sample standard deviation
is 17.6 seconds. Waiting times are known to follow a normal distribution.
at the 1% level of significance.
1. Let a=0.01
2. Set up hypothesis:
a. 𝐻0 : μ = 60 vs. 𝐻𝑎 : μ < 60
*Notice that we phrase the hypotheses in terms of seconds, since our data were
measured in seconds.
3. Rejection Rule:
a. We will reject 𝐻0 if our P-value is < 0.01.
4. The test statistic is:
a. The P-value is P(T(30)≤- 1.61) = P(T(30)≥ -1.61) by the symmetry of the t
distribution. We see from Table D that P(T(30) ≥1.310) = 0.10 and P(T(30)
≥1.697) = 0.05. Since 1.310 < 1.61 < 1.697, it follows that the P-value is between
0.05 and 0.10. For any P-value between 0.05 and 0.10, we fail to reject the null
hypothesis, so our conclusion is as follows£¡
Conclusion: Since the P-value > 0.01, we fail to reject the null hypothesis. At the 1% level
of significance, we have insufficient evidence to conclude that the true mean waiting
time in the drive thru is less than one minute.
Example- Grooms:
In 1968, census results indicated that the age at which American men first married had a
mean of 23.3 years.
We take a random sample of 28 men who married for the first time in 2008. Their
average age was 25.2 years and the standard deviation of their ages was 4.6 years. We
will assume for the purpose of this example that ages of first-time grooms are normally
distributed.
What is a 95% confidence interval for the true mean age of first-time grooms in 2008?
𝑥±𝑡𝑛−1
∗𝑠
√𝑛
25.2 ±(2.052)(4.6/√28)
= (23.42, 26.98)
We used the upper 0.025 critical value from the t distribution with n 1 = 27 degrees of
freedom.
Cautions About t procedures
Because t is a function of the sample mean and of the sample standard deviation, which
are both strongly influenced by outliers, t itself is strongly influenced by outliers.
Although we have an assumption that the population is normal, the t procedures are
quite robust against non normality.
The most important assumption (except for small samples) is that the data are from an
SRS from the population.
For large sample sizes, the violation of the normality assumption is not a serious one.
Paired Data
We will often encounter a situation where data are collected in pairs. In such situations,
we will usually be interested not in the individual observations, but rather in the
differences in values of some variable for each pair.
The term paired data means that the data have been observed in natural pairs. There
are several ways in which paired data can occur:
Two different variables are measured for each individual. We are interested in
the difference between the values of the two variables.
Document Summary
A fast food restaurant claims that the average waiting time in their drive through is less than a minute. We record the drive thru waiting times for a random sample of 31 customers. The sample average time is 54. 9 seconds and the sample standard deviation is 17. 6 seconds. Waiting times are known to follow a normal distribution. We will co(cid:374)duct a h(cid:455)pothesis test to e(cid:454)a(cid:373)i(cid:374)e the sig(cid:374)ifica(cid:374)ce of the restaura(cid:374)t"s clai(cid:373) at the 1% level of significance: let a=0. 01, set up hypothesis: measured in seconds. We see from table d that p(t(30) (cid:3410)1. 310) = 0. 10 and p(t(30) (cid:3410)1. 697) = 0. 05. Since 1. 310 < 1. 61 < 1. 697, it follows that the p-value is between. For any p-value between 0. 05 and 0. 10, we fail to reject the null hypothesis, so our conclusion is as followsi : rejection rule, the test statistic is: Conclusion: since the p-value > 0. 01, we fail to reject the null hypothesis.
