Log-in | Contact Jeff | Email Updates

Question 466:

1

Answer:

No answer provided yet.
  1. To construct the 90% confidence interval we need to find the margin of error. The margin of error is the standard error of the mean (SEM) times a critical value from the normal distribution. The value for a 90% confidence level can be looked-up in a normal table or from the percentile to z-score calculator. We get a value of 1.644 when we enter .90 2-sided in the calculator. Now, use the following steps to construct the standard error of the mean.

     

    1.      First we find the proportion of unpopped kernels, which is 86/773 = .1112 and is denoted p.

    2.      The standard error of the mean (SEM) is the standard deviation divided by the square root of the sample size.

    3.      The standard deviation is SQRT(pq), which is p (.1112) times q, which is 1-p (.8887). So SQRT(.1112*.8887) = .3144

    4.      The square root of the sample size = SQRT(773) = 27.803 making the SEM .3144/27.803 = .01131

    5.      The margin of error is then .01131*1.644 = .0186.

    6.      The confidence interval is found by adding and subtracting the margin of error to the proportion. .1112-.0186 and .1112+.0186 provides us with a 90% confidence interval around the unpopped kernels of between .0926 and .1298

     

    The normality assumption refers to the normal approximation to the binomial and means its reasonable to use the z-score (critical value) from the normal distribution as long as our sample is sufficiently large and not too close to either 1 or 0. One general rule of thumb is to have np >5 and nq >5. So that's 773*.1112 = 86 and 773*.8887 = 687 so we could say the normality assumption is reasonable here considering both are well above 5.

     

    The very Quick rule is not a common statistical term or technique but it is my understanding that this refers to a short-cut in making estimates of the confidence interval when the proportion is near .5 and when using a 95% confidence level. Since we're using a 90% Confidence level and the proportion is not near .5, it would not apply.

     

    The reason this sample might not be typical is because it represents data from only 1 bag and we are making inferences about all pop-corn kernels based on just one-bag, experimentally it is more sound to sample multiple bags, the compute the confidence interval on the mean number of non-popped kernels.

We can confirm our results by entering 86 in passed and 773 in total tested on the confidence interval around a proportion calculator. Select the 90% confidence interval and look at the "Wald" results.

Not what you were looking for or need help?

Ask a new Question

Browse All 869 Questions

Search All Questions: