Test the hypothesis that the average content of containers of a particular lubricant is 15 liters if the contents of a random sample of 15 containers are 10.2, 9.7, 10.1, 10.3, 10.1, 10.3, 10.1, 9.8, 9.9, 10.4, 10.3, 10.5,11.2,11.3 and 9.8 liters. Use a 0.01 level of significance and assume that the distribution of contents is normal.
Also find 95% confidence interval for the population mean. Interpret your results

Answer:

Kindly check explanation

Step-by-step explanation:

Given The samples:

X = 10.2, 9.7, 10.1, 10.3, 10.1, 10.3, 10.1, 9.8, 9.9, 10.4, 10.3, 10.5,11.2,11.3, 9.8

Using a calculator ;

Sample mean, xbar = 10.27

Sample standard deviation, s = 0.462

H0 : μ = 15

H1 : μ < 15

The test statistic :

(xbar - μ) ÷ (s/sqrt(n))

(10.27 - 15) / (0.462/sqrt(15))

-4.73 / 0.1192878

= - 39.65

Using the Pvalue form Tscore, calculator, df = 14

Pvalue = 0.000001,

Since, Pvalue < α ; then we reject the Null

Confidence interval :

Xbar ± Margin of error

Margin of Error = Tcritical * s/√n

TCritical at 95%, df = 14

Margin of Error = 2.145 * 0.462/sqrt(15) = 0.2559

Lower boundary :

10.27 - 0.2559 = 10.0141

Upper boundary :

10.27 + 0.2559 = 10.5259

(10.0141 ; 10.5259)

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