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Hey everyone, welcome to numerate.
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We were given a problem here by a survey conducted by industry in the computer software around and given a mean number of six days, 9 .4 and a standard deviation 2 .7.
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We have a sample size 15 with the list of six days given, and we were asked to run a test.
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To see if the sick days is actually less than 9 .4 for the mean at a 0 .05 significance level.
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So from here, we will write the null hypothesis first, and we will do a z test since we are given the population standard deviation.
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Okay, so no hypothesis, since we were looking for fewer than 9 .4, we would do a mule for the average is less than 9 .4.
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And then from here, we would do the alternative hypothesis, where mu is greater than 9 .4, which is the opposite.
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And let's dive into our z test.
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So our z score equation is an x, which is the population mean, minus the mule, which is the sample mean, divided by standard deviation, divided by the zero.
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By square root n.
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And from here, let's define our variables, write out our variable values.
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So this usually is x bar equals.
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So x bar, in this case, what the population mean.
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So it's going to be 9 .4.
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And in meal, mu will equal.
02:00
So we had to sum up all the values in the list and it divided by the number, the number would be 15.
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So x divided by the number.
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The number is 15, so let's just plug you down for 15...