Bottles of a popular cola are supposed to contain 400 milliliters (ml) of cola. There is some variation from bottle to bottle because the filling machinery is not perfectly precise. The distribution of the contents is Normal with standard deviation 3 ml. An inspector suspects that the bottle is under filling. He measures the contents of six bottles. The results are 399.4, 397.5, 401.5, 398.9, 400.5, 401.0. Is this convincing evidence that the mean content of cola bottles is less than the advertised 400 ml? I. State the hypothesis that you will test. II. Calculate the test statistic. III. Find the P-value. IV. Do we reject the null hypothesis at the significance level 5%? State your conclusion.

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Bottles of a popular cola are supposed to contain 400 milliliters (ml) of cola. There is some variation from bottle to bottle because the filling machinery is not perfectly precise. The distribution of the contents is Normal with standard deviation 3 ml. An inspector suspects that the bottle is under filling. He measures the contents of six bottles. The results are 399.4, 397.5, 401.5, 398.9, 400.5, 401.0. Is this convincing evidence that the mean content of cola bottles is less than the advertised 400 ml?                          I.       State the hypothesis that you will test.                        II.          Calculate the test statistic.                      III.          Find the P-value.                      IV.           Do we reject the null hypothesis at the significance level 5%? State your conclusion.

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The mean is the sum of all values divided by the number of values:x―=399.4+397.5+401.5+398.9+400.5+401.06≈399.8n is the number of vlaues in the sampleThe standard deviation is the square root of the sum of squared deviations from the mean divided by n−1s―=(399.4−399.8)2+(397.5−399.8)2+(401.5−399.8)2+(398.9−399.8)2+(400.5−399.8)2+(401.0−399.8)26−1≈1.48862Determine the hypothesis::H0:μ=400Ha:μ≠400Determine the values of the test statistic:t=x―−μ0s/n=399.8−4001.48862/6=−0,329The P-value is the probability of obtaining the value of the test statistic, or a value more extreme. The P-value is the number (or interval) in the column title of Table B containing the t-value in the row df=n−1=20−1=19:0.1=2×0.05&amp;lt;P&amp;lt;2×0.10=0.20If the P-value is smaller than the significance level, then the null hypothesis is rejected.P&amp;gt;0.05=5%⇒ Fail to reject H0There is not sufficient evidence to support the claim.

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