Question

Industrial wastes and sewage dumped into our rivers and streams absorb oxygen and thereby reduce the amount of dissolved oxygen available for fish and other forms of aquatic life. One state agency requires a minimum of 5 parts per million (ppm) of dissolved oxygen in order for the oxygen content to be sufficient to support aquatic life. Six water specimens taken from a river at a specific location during the low-water season (July) gave readings of 4.9, 5.0, 5.0, 5.0, 5.0, and 4.6 ppm of dissolved oxygen. Do the data provide sufficient evidence to indicate that the dissolved oxygen content is less than 5 ppm? Test using α = 0.05.

H₀: μ = 5 versus H_a: μ < 5    

State the test statistic. (Round your answer to three decimal places.)

State the rejection region. (If the test is one-tailed, enter NONE for the unused region. Round your answers to three decimal places.)

Answer 1

To conduct the hypothesis test for the given claim and the given sample 4.9, 5.0, 5.0, 5.0, 5.0, and 4.6 ppm we need to find the sample mean and sample standard deviation which is calculated as:

Mean = (4.9 + 5.0 + 5.0 + 5.0 + 5.0 + 4.6)/6
= 29.5/6
Mean = 4.9167

and the sample standard deviation as:

Standard Deviation s= √(1/6 - 1) x ((4.9 - 4.9167)2 + ( 5.0 - 4.9167)2 + ( 5.0 - 4.9167)2 + ( 5.0 - 4.9167)2 + ( 5.0 - 4.9167)2 + ( 4.6 - 4.9167)2)
= √(1/5) x ((-0.016699999999999)2 + (0.0833)2 + (0.0833)2 + (0.0833)2 + (0.0833)2 + (-0.3167)2)
= √(0.2) x ((0.00027888999999998) + (0.0069388900000001) + (0.0069388900000001) + (0.0069388900000001) + (0.0069388900000001) + (0.10029889))
= √(0.2) x (0.12833334)
= √(0.025666668)
= 0.1602

Now the hypotheses are:

Rejection region:

Based on the goven confidence level that is 1-0.05 -0.95 and the degree of freedom which is calculated as df = n-1= 6-1= 5, the critical tc is computed using excel formula for T-distribution which is =T.INV(0.05,5)

Thus tc computed as −2.015. So the rejection region is reject Ho if t <tc.

Test statistic:

P-value:

Based on the test statistic and degree of freedom the P-value is computed using excel formula for normal distribution which is =T.DIST(-1.274,5,TRUE), thus the p-value computed as 0.1294.

Conclusion:

Since P-value is greater than 0.05 hence we cannot reject the null hypothesis and conclude that there is insufficeint evidence to support the claim.