Question

When σ (population standard deviation) is unknown [Critical Value Approach]

The value of the test statistic t for the sample mean is computed as follows:

      t =   with degrees of freedom d.f. = n – 1

where n is the sample size, µ is the mean of the null hypothesis, H0, and s is the standard deviation of the sample.

      t is in the rejection zone: Reject the null hypothesis, H0

      t is not in the rejection zone: Do not reject the null hypothesis

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1. The Environmental Protection Agency (EPA) recommends that the sodium content in public water supplies should be no more than 20 mg per liter. Forty-nine samples were taken at random from water taps in Bethlehem. The sample average was = 21.8 mg per liter. The sample standard deviation was s = 5.6 mg per liter. Does the average sodium content in the water supply in Bethlehem exceed the recommended level?

Do hypothesis test for = 0.01 and for = 0.05

For the null hypothesis, assume that the water quality satisfies the EPA requirement of no more than 20 mg per liter.

(a) State the null and the alternative hypothesis.

      H0 : µ =

         Ha : µ [ ≠ or > or < ]

(b) Should the test be a left-tail, right-tail or two- tail test?

(c) Calculate t using the formula above.

(d) Find the rejection region (critical value) for = 0.01. What is the decision? [Use the t table and the values for significance in the row for one-tail α. Values for t will be in the row for n-1 degrees of freedom.]

Is the null hypothesis rejected or does the water quality satisfy EPA standards?

(e) Find the rejection region (critical value) for = 0.05. What is the decision? [Use the t table and the values for significance in the row for one-tail α. Values for t will be in the row for n-1 degrees of freedom.]

Is the null hypothesis rejected or does the water quality satisfy EPA standards?

Answer 1

a) Here we have to test that

Null hypothesis :

Alternative hypothesis :

where

b) Test is right tailed test.

c)

n = sample size = 49

Sample mean =

Sample standard deviation = s = 5.6

Here population standard deviation is not known so we use t test.

Test statistic :

t = 2.25

Test statistic = 2.25

d) Level of significance = = 0.01

Degrees of freedom = n - 1 = 49 - 1 = 48

There is no approimate value for df = 48 in statistical table so we use excel.

Critical value for 1 - = 1- 0.01= 0.99 and df = 48 from excel using function :

=T.INV(0.99,48) (Because T.INV gives us left tail value so we do 1 - 0.01 = 0.99)

= 2.407 (Round to 3 decimal)

Critical value = 2.407

Test is right tailed test.

Here test statistic < Critical value

So we fail to reject H0.

Conclusion : There is not sufficient evidence to conclude that the average sodium content in the water supply in Bethlehem exceeds the recommended level.

e) Level of significance = = 0.05

Degrees of freedom = n - 1 = 49 - 1 = 48

There is no approimate value for df = 48 in statistical table so we use excel.

Critical value for 1 - = 1- 0.05= 0.95 and df = 48 from excel using function :

=T.INV(0.95,48) (Because T.INV gives us left tail value so we do 1 - 0.01 = 0.95)

= 1.677 (Round to 3 decimal)

Critical value = 1.677

Test is right tailed test.

Here test statistic > Critical value

So we reject H0.

Conclusion : There is sufficient evidence to conclude that the average sodium content in the water supply in Bethlehem exceeds the recommended level.