A machine that is programmed to package 1.95 pounds of cereal in each cereal box

Question

A machine that is programmed to package 1.95 pounds of cereal in each cereal box is being tested for its accuracy. In a sample of 26 cereal boxes, the mean and standard deviation are calculated as 1.99 pounds and 0.15 pound, respectively. Use Table 2. a. Select the null and the alternative hypotheses to determine if the machine is working improperly, that is, it is either underfilling or overfilling the cereal boxes. H0: µ 1.95; HA: µ < 1.95 H0: µ 1.95; HA: µ > 1.95 H0: µ = 1.95; HA: µ 1.95 b. Calculate the value of the test statistic. (Round your answer to 2 decimal places.) Test statistic c-1. Approximate the p-value. 0.10< p-value < 0.20 0.01 < p-value < 0.02 0.02 < p-value < 0.05 p-value < 0.01 p-value Picture 0.2 c-2. What is the conclusion at the 10% significance level? Reject H0 since the p-value is greater than . Reject H0 since the p-value is smaller than . Do not reject H0 since the p-value is greater than . Do not reject H0 since the p-value is smaller than . d-1. Calculate the critical value(s) at a 10% level of significance. (Round your answer to 3 decimal places.) Critical value(s) ± d-2. Can you conclude that the machine is working improperly? Yes No

Explanation / Answer

1) µ = 1.95; HA: µ 1.95

b) std error of mean =std deviation/(n)1/2 =0.15/(26)1/2 =0.0294

test statistic t=(X-mean)/std error =(1.99-1.95)/0.0294=1.3597

c-1)p-value. 0.10< p-value < 0.20

c-2) Do not reject H0 since the p-value is greater than

d-1) critical value(s) at a 10% level of significance +/- 1.708

as we can not reject null hyothessis; therefore No

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