Please help me with this question
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I have an expression
floating around in my head; let's see if it makes sense.
The variance of binary valued random variable b that comes up 1 with probability p (so has mean p) is
That's for an individual sample. For the observed average we divide by n, and for the standard deviation we take the square root:
Plugging in the numbers,
One standard deviation of the average is almost 2% so a 27% outcome was 3/1.9 = 1.6 standard deviations from the mean, corresponding to a two sided probability of a bit bigger than 10% of happening by chance.
So this is borderline suspect; most surveys will include a two sigma margin of error, say plus or minus 4 percent here, and the results were within those bounds.
The valid conclusions for the manager based on the considered test is given by: Option
<h3>When do we perform one sample z-test?</h3>One sample z-test is performed if the sample size is large enough (n > 30) and we want to know if the sample comes from the specific population.
For this case, we're specified that:
- Population mean =
= $150
- Population standard deviation =
= $30.20
- Sample mean =
= $160
- Sample size = n = 40 > 30
- Level of significance =
= 2.5% = 0.025
- We want to determine if the average customer spends more in his store than the national average.
Forming hypotheses:
- Null Hypothesis: Nullifies what we're trying to determine. Assumes that the average customer doesn't spend more in the store than the national average. Symbolically, we get:
- Alternate hypothesis: Assumes that customer spends more in his store than the national average. Symbolically
where is the hypothesized population mean of the money his customer spends in his store.
The z-test statistic we get is:
The test is single tailed, (right tailed).
The critical value of z at level of significance 0.025 is 1.96
Since we've got 2.904 > 1.96, so we reject the null hypothesis.
(as for right tailed test, we reject null hypothesis if the test statistic is > critical value).
Thus, we accept the alternate hypothesis that customer spends more in his store than the national average.
Learn more about one-sample z-test here: