A simple random sample was taken of 44 water bottles from a bottling plant’s warehouse. The dissolved oxygen content (in mg/L) was measured for each bottle, with these results: 11.53, 8.35, 11.66, 11.54, 9.83, 5.92, 7.14, 8.41, 8.99, 13.81, 10.53, 7.4, 6.7, 8.42, 8.4, 8.18, 9.5, 7.22, 9.87, 6.52, 8.55, 9.75, 9.27, 10.61, 8.89, 10.01, 11.17, 7.62, 6.43, 9.09, 8.53, 7.91, 8.13, 7.7, 10.45, 11.3, 10.98, 8.14, 11.48, 8.44, 12.52, 10.12, 8.09, 7.34 Here the sample mean is 9.14 mg/L. The population standard deviation of the dissolved oxygen content for the warehouse is known from long experience to be about σ = 2 mg/L. Consider testing H0 : µ = 10 vs. HA : µ 6= 10, where µ is the unknown population mean dissolved oxygen content, at significance level α = .02 (not the usual .05). (a) A Z test is appropriate here because we have a SRS with a large n. Find the value of the test statistic and the p-value. (b) What conclusion do you reach?

(a) Test statistic is -2.85 and p-value is 0.0022

(b) Reject the null hypothesis. The population mean of dissolved oxygen content is not equal to 10 mg/L

Step-by-step explanation:

H0: mu equals 10

Ha: mu not equals 10

The test is a two-tailed test because the alternate hypothesis is expressed using not equal to

(a) Test statistic (z) = (sample mean - population mean) ÷ (sd/√n) = (9.14 - 10) ÷ (2/√44) = -0.86 ÷ 0.302 = -2.85

Cumulative area of the test statistic = 0.9978

p-value = 2(1 - 0.9978) = 2(0.0022) = 0.0044

(b) The critical value using 0.02 significance level is 2.422. For a two-tailed test, the region of no rejection of the test statistic lies between -2.422 and 2.422.

Conclusion:

Reject the null hypothesis because the test statistic -2.85 falls outside the region bounded by the critical values -2.422 and 2.422.

The population mean of dissolved oxygen content is not equal to 10 mg/L

y= 2x+7

changed to y= 5x +7

as slope is increased from 2 to 5

therefore it would be less steeper

so, option d) is correct