Write 6/5 as a percent?
2 answers:
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Answer:
There is enough evidence to say that the true average heat output of persons with the syndrmoe differs from the true average heat output of non-sufferers.
Step-by-step explanation:
We have to perform a hypothesis test on the difference between means.
The null and alternative hypothesis are:
μ1: mean heat output for subjects with the syndrome.
μ2: mean heat output for non-sufferers.
We will use a significance level of 0.05.
The difference between sample means is:
The standard error is
The t-statistic is
The degrees of freedom are
The critical value for a left tailed test at a significance level of 0.05 and 16 degrees of freedom is t=-1.746.
The t-statistic is below the critical value, so it lies in the rejection region.
The null hypothesis is rejected.
There is enough evidence to say that the true average heat output of persons with the syndrmoe differs from the true average heat output of non-sufferers.