Problem 1:

Farmasi UMS Semester Genap 2012/2013

Tutorial: Hypothesis Testing

Conceptual Questions 1. In hypothesis testing, the hypothesis tentatively assumed to be true is

a.the alternative hypothesis

b.the null hypothesis

c.either the null or the alternative

d.All of the above answers are correct.

e.None of the above answers is correct.

ANSWER:b

2. In hypothesis testing if the null hypothesis is rejected,

a.no conclusions can be drawn from the test

b.the alternative hypothesis must also be rejected

c.the data must have been accumulated incorrectly

d.the sample size has been too small

e.None of the above answers is correct.

ANSWER:e

3. The level of significance is the

a.maximum allowable probability of Type II error

b.maximum allowable probability of Type I error

c.same as the confidence coefficient

d.same as the p-value

e.None of the above answers is correct.

ANSWER:b

4. The level of significance in hypothesis testing is the probability of

a.accepting a true null hypothesis

b.accepting a false null hypothesis

c.rejecting a true null hypothesis

d.could be any of the above, depending on the situation

e.None of the above answers is correct.

ANSWER:c

5. The probability of making a Type I error is denoted by

a.(

b.(

c.1 - (

d.1 - (

e.None of the above answers is correct.

ANSWER:a

6. When the following hypotheses are being tested at a level of significance of (

H0: ( ( 100

Ha: ( < 100

the null hypothesis will be rejected if the test statistic Z is

a.> Z(

b.> Z(

c.< -Z(

d.< 100

e.None of the above answers is correct.

ANSWER:c

7. When the p-value is used for hypothesis testing, the null hypothesis is rejected if

a.p-value < (

b.( < P-value

c.p-value > (

d.p-value = (

e.None of the above answers is correct.

ANSWER:a

8. In order to test the following hypotheses at ( level of significance

H0: ( ( 100

Ha: ( > 100

the null hypothesis will be rejected if the test statistic Z is

a.> Z(

b.< Z(

c.< -Z(

d.< 100

e.None of the above answers is correct.

ANSWER:a

9. Which of the following does not need to be known in order to compute the p-value?

a.knowledge of whether the test is one-tailed or two-tailed

b.the value of the test statistic

c.the level of significance

d.All of the above are needed.

e.None of the above answers is correct.

ANSWER:c

10. In the hypothesis testing procedure, ( is

a.the level of significance

b.the critical value

c.the confidence level

d.1 - level of significance

e.None of the above answers is correct.

ANSWER:a

11. A soft drink filling machine, when in perfect adjustment, fills the bottles with 12 ounces of soft drink. Any overfilling or under filling results in the shutdown and readjustment of the machine. To determine whether or not the machine is properly adjusted, the correct set of hypotheses is

a.H0: ( < 12

Ha: ( ( 12

b.H0: ( ( 12

Ha: ( > 12

c.H0: ( ( 12

Ha: ( = 12

d.H0: ( = 12

Ha: ( ( 12

e.None of the above answers is correct.

ANSWER:d

Problem 1: The Naphtha reformer unit in a refinery produces an octane number of 90 under an existing (old) catalyst. A new catalyst has been developed and used in the unit. An average octane number of 92 has been estimated using a sample of 4 data points using this new catalyst. Has the octane number changed significantly? Use a hypothesis test at the 5% significance levelFrom past experience, we know that the standard deviation for octane number (in octane units) is 1.5 and we are willing to assume that the variance hasnt changed. Solution:

Hypothesis test - 1. Null hypothesis

H0: ( = 90

2. Alternate hypothesis

Ha: ( ( 90

3. Test statistic:

with substitution of the sample average=92, n=4 and (=1.5 for our example gives z =2.67

4. Rejection criteria:

Is z = 2.67 inside the range of values for Z that contains the middle 95% of the probability?

NO, z = 2.67 isnt between -1.96 and +1.96. (See yellow table for rejection criteria).Therefore, it doesnt seem to make sense that ( = 90.

We say that we reject the null hypothesis and we conclude, at the 95% confidence level, that ( ( 90 using the new catalyst.

Problem 2: The yield of a chemical reactor is being studied. Previous experience has indicated that the standard deviation of yield for the reactor is 3%. The past five days of plant operation have provided yields of 91.6%, 88.75%, 90.8%, 89.95% and 91.3%. (a) The previous yield was 90%. Has the yield of this reactor changed under current operation? Test at the 5% significance level.

(b) What is the type II error probability for this experiment if the true mean yield is 92%?

Solution:

In this question, we are studying chemical reactor yield. We are given a standard deviation value of 3 -- the language in the question indicates that this is considered to be a KNOWN value (it is based on extensive previous operating experience, and no indication is given of the number of data points from which it might have been computed). Thus, we should be using hypothesis tests or confidence intervals for means, with KNOWN variance.

The average yield computed from the data is 90.48% .

Hypothesis test - a 2-sided test is required ("Has the yield of this reactor changed under current operation?").

This is a test of a value of the mean, with known variance.

H0: mean = 90

Ha: mean not equal to 90

The test statistic is

since sigma is 3, and n=5. This absolute value of this test statistic is compared against z0.025 = 1.96, and the null hypothesis is not rejected, i.e., we conclude that the yield has not changed.

Confidence interval approach

The confidence interval is

which produces (87.85, 93.11). Since this interval contains 95%, we conclude that the yield has not changed.

(b) What is the type II error probability for this experiment if the true mean yield is 92%?

The acceptance region for hypothesis in part (a) is

Note here, the acceptance region is a region centred at 90, not the confidence interval that we got in part (a).

In order to calculate the type II error, we need to calculate the area under the normal curve with mean 92 and standard deviation 3, between the interval (87.37, 92.63).The new statistic is

Convert the interval (87.37, 92.63) to standard normal fence values as

and

Then the type II error is

This is the same if we calculate a shift in the original standard normal fence values and . This shift is

So the new fence values of standard normal distribution is

The new fence values are the same as what we got above, so the type II error is

Problem 3: The sugar content in the syrup of canned peaches has been found to be Normally distributed. The variance is thought to be 18 mg2. A random sample of 10 cans has been selected, and has yielded a sample standard deviation of 4.8 mg. Does this observation support the conclusion that the population variance is 18 mg2?

Use a hypothesis test at the 5% significance level.Solution:

We are interested in testing whether the sugar content variance is 18 mg2. We can assess this using a hypothesis test on variance. The estimated variance is 4.82 = 23.04. The hypotheses are as follows:

H0: variance = nominal variance of 18

Ha: variance not equal to nominal variance of 18

The test statistic is:

Substituting in the values, we obtain

9(23.04)/18 = 11.52

We need to compare this to two fences with an outer tail area of 5% (I have decided to test at the 5% significance level). The values are

Since the test statistic lies in between these two fences, we do not reject the null hypothesis, and conclude that the sugar content is plausibly 18 mg2.

Confidence interval approach

The appropriate form of the confidence interval is:

The chi-squared values are 2.70 and 19 for the upper tail areas of 0.975 and 0.025 respectively. The confidence interval is (10.91, 76.8). Since this interval contains 18, we conclude that it is plausible that the variance of the syrup sugar concentration is 18.

Problem 4: The wall thickness of 25 glass 2 L bottles is being monitored by a quality control engineer. The average of the bottles has been computed to be 4.05 mm, and the sample standard deviation of the bottles is 0.08 mm. In order to confirm that quality specifications are being met, we must demonstrate that the mean wall thickness is greater than 4 mm. Use a hypothesis test at the 5% significance level to assess whether this condition is being met.

Solution:

In order to assess whether the wall thickness is greater than 4 mm, use the following null and alternate hypotheses:

H0: mean = 4 mm

Ha: mean > 4 mm

The test statistic in this case is

because the nominal value of the mean is 4 mm, we have estimated the average using 25 data points, and we are estimating the variance from the data.

We now have to compare this test statistic against the value from the Students t-distribution with 24 degrees of freedom (n-1=25-1=24), and an upper tail area of 5% (since this is a one-sided test). The corresponding t-value is 1.711. Since the test statistic value is greater than the fence, we reject the null hypothesis, and conclude that the wall thickness is greater than 4 mm.

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