Title: Testing the Mean
1Testing the Mean
2- To test µ when s is known
- State the null and alternate hypothesis and set
the level of significance a - If you can assume that x has a normal
distribution, then any sample size will work. If
you cannot, then use a sample - size 30
- Use information to calculate E using the z value
or the TI-83 with a Z-test - If P-value a reject H0. If P-value gt a, then
do not reject H0 - State your conclusion in the context of the
application.
3(No Transcript)
4- H0 µ 41 and H1 µ gt 41
- Compute E using Z test, s 35, n 40 and
47. - P .1401 gt a .05
- So, We do not reject H0.
- Based on this test we conclude that the sunspot
activity - during the Spanish Colonial Period was
higher than the - long-term mean
5- To test µ when s is unknown
- Take a sample of size n and calculate the sample
mean and the sample standard deviation sx. - State the null and alternate hypothesis and set
the level of significance a - If you can assume that x has a normal
distribution, then any sample size will work. If
you cannot, then use a sample - size 30
- Use information to calculate P using the t value
or the TI-83 with a t-test (Student's t test) - If P-value a reject H0. If P-value gt a, then
do not reject H0 - State your conclusion in the context of the
application.
6Hypothesis Testing of the Mean Whens is Unknown
Using the Ti-83
- Press the STAT button
- Move the cursor to TESTS and Select 2T-Test
- Press Enter
- Highlight Stats
- Enter requested information µ0, Sx, x, n
- Highlight the appropriate alternative hypothesis
- µ? µ0 (Two Tailed), µgt µ0 (Right Tailed),
µltµ0 (Left Tailed) - Highlight Calculate and press ENTER
- Read t value and p probability of H0
- If p ? we reject H0 and if p gt ? we do not
reject H0
7Using TI-83 with t test, we calculate P .0478 gt
a .01 We therefore, do not reject H0
8So, based on the data, we cannot say that the
drug 6-mP provides a different average remission
time than the previous drug.
9Use a 10 level of significance to test the claim
that the mean weight of fish caught in a lake is
2.1 kg (against the alternate that the weight is
lower). A sample of five fish weighed an
average of 1.99 kg with a standard deviation of
0.09 kg.
10 test the claim that the mean weight of fish
caught in a lake is 2.1 kg (against the alternate
that the weight is lower). ...
- H0 ? 2.1
- H1 ? lt 2.1
- Enter Data In TI-83
- Calculate p .02614 lt .1 (10)
- We reject Ho
11We conclude (at 10 level of significance) that
the true weight of the fish in the lake is less
than 2.1 kg.
12H0 µ 41.7 H1 µ 41.7, a
.05 Using Z-Test (We know s) P .046 a
.05 We reject H0 At 5 significance there
is enough evidence to conclude that with the new
e-mail system the number of E-mails is different.