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Provided with information on a sample (>30) with sample mean and standard deviation, use the z test to either accept or reject a given null hypothesis.

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A leading manufacturer claims its product contains {thisamount} ml by volume. A sample of {n} items was taken and the mean of the sample was $\\bar{x}=\\var{m}$ ml and the standard deviation of the sample was $s = \\var{stand}$ ml. Is the manufacturers claim justified?

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a)

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Step 1: Null Hypothesis

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$\\operatorname{H}_0\\;: \\; \\mu=\\;\\var{thisamount}$

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Step 2: Alternative Hypothesis

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$\\operatorname{H}_1\\;: \\; \\mu \\neq\\;\\var{thisamount}$

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b)

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We should use the z statistic since the sample size is greater than 30.

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The test statistic:

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\\[z = \\frac{\\var{m}-\\var{thisamount}}{\\frac{\\var{stand}}{\\sqrt{\\var{n}}}}=\\var{teststatistic}\\]

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to 3 decimal places.

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c)

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 For a significance level of 5%:

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$z_{{\\frac{\\alpha}{2}}}=\\var{critical}$

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d)

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Hence there is {evi1[pval]} evidence against $\\operatorname{H}_0$ and so we {dothis} $\\operatorname{H}_0$.

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{Correctc}

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Step 1: Null Hypothesis

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$\\operatorname{H}_0\\;: \\; \\mu=\\;$[[0]]

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Step 2: Alternative Hypothesis

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$\\operatorname{H}_1\\;: \\; \\mu \\neq\\;$[[1]]

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Step 3: Test statistic

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Should we use the z or t test statistic? [[0]] (enter z or t).

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Now calculate the test statistic = ? [[1]] (to 3 decimal places)

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Step 4: critical value

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Use tables to find the critical value for a significance level of 5%. 

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[[0]]

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Your Decision:

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[[0]]

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Conclusion:

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[[1]]

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