Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

In a study of larval development in the tufted apple budmoth (Platynota idaeusalis), an entomologist measured the head widths of 50 larvae. All 50 larvae had been reared under identical conditions and had moulted six times. The mean head width was \(1.20 \mathrm{~mm}\) and the standard deviation was \(0.14 \mathrm{~mm}\). Construct a \(90 \%\) confidence interval for the population mean. \(^{17}\)

Short Answer

Expert verified
The 90% confidence interval for the population mean is (mean head width - t* x SE) to (mean head width + t* x SE), where mean head width is 1.20 mm, t* is the t-critical value for df=49 and 90% confidence level, and SE is the standard error calculated from the sample standard deviation and size.

Step by step solution

01

Determine the appropriate statistical test

Since we are constructing a confidence interval for the population mean based on a sample, we need to use a t-distribution if the population standard deviation is not known. Here, we will use the sample standard deviation and thus apply the t-distribution, as the sample size is relatively small (n=50) and the population standard deviation is not given.
02

Find the degrees of freedom

The degrees of freedom (df) for a t-distribution are found by subtracting 1 from the sample size (n). In this case, the degrees of freedom is 50 - 1 = 49.
03

Determine the t-critical value

For a 90% confidence level and 49 degrees of freedom, refer to the t-distribution table or use a calculator to find the t-critical value, which we will denote as t*. This is the value that the sample mean needs to exceed to be in the top 5% of a t-distribution (since 5% is left in the two tails combined for a 90% confidence interval).
04

Calculate the standard error

The standard error (SE) is the standard deviation divided by the square root of the sample size: SE = s / √n. Here, the standard deviation s is 0.14 mm and n=50.
05

Construct the confidence interval

The confidence interval is the range from (sample mean - t* x SE) to (sample mean + t* x SE). Plug in the mean head width, the t-critical value, and the standard error to calculate the interval.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the T-Distribution
The t-distribution, also known as the Student's t-distribution, is a probability distribution that is symmetrical and bell-shaped, much like the standard normal distribution. However, it has heavier tails, meaning it tends to produce values that fall further away from the mean, which is important when dealing with small sample sizes.

This comes into play especially when we do not know the true population standard deviation and instead have to rely on sample data. The t-distribution accounts for the additional uncertainty that comes with estimating the population standard deviation by the sample standard deviation. As the sample size increases, the t-distribution approaches the normal distribution. The t-distribution is critical in the construction of confidence intervals and hypothesis tests for population means when the sample size is small and the population standard deviation is unknown.
Degrees of Freedom Explained
Degrees of freedom, often abbreviated as df, are a measure of the amount of independent information available to estimate a parameter. In the context of the t-distribution, the degrees of freedom are equivalent to the number of independent observations minus the number of parameters being estimated.

For example, when estimating a population mean from a sample, the degrees of freedom are calculated as the sample size (n) minus one (df = n - 1). This subtraction accounts for the fact that we're using the sample mean to estimate one population parameter. The number of degrees of freedom affects the shape of the t-distribution; the larger the degrees of freedom, the closer the distribution will be to the normal distribution. In the scenario of the tufted apple budmoth larvae, with 50 larvae measured, we have 49 degrees of freedom.
Standard Error and its Significance
Standard error (SE) measures the precision of a sample mean as an estimator of the population mean. It quantifies how much the sample means you would expect to vary from the actual population mean if you were to take multiple samples from the population.

The formula for calculating the standard error is SE = s / √n, where 's' represents the sample standard deviation and 'n' is the sample size. A smaller standard, error then, indicates a more precise estimate of the population mean. It's crucial for determining the margin of error in a confidence interval; a narrower interval suggests a more accurate estimation. Given a standard deviation of 0.14 mm for the head widths of the larvae and a sample size of 50, one would compute the standard error to further facilitate the confidence interval calculation.
Population Mean Estimation
Estimating the population mean involves using sample data to infer the mean of the entire population from which the sample was taken. The point estimate for the population mean is the sample mean, but due to the uncertainty inherent in sampling, we also provide a range of values in which we are fairly confident the true population mean lies; this range is known as a confidence interval.

A confidence interval spans from the lower to the upper bound and is calculated using the formula (sample mean ± t* × SE), where 't*' is the t-critical value from the t-distribution for the specified confidence level and degrees of freedom. This interval estimates where the true population mean is likely to be, with a certain degree of confidence. In our exercise related to the tufted apple budmoth, we seek to construct a 90% confidence interval, which means we can say that, based on our sample, there is a 90% chance that the population mean head width falls within our calculated range.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose you are planning an experiment to test the effects of various diets on the weight gain of young turkeys. The observed variable will be \(Y=\) weight gain in 3 weeks (measured over a period starting 1 week after hatching and ending 3 weeks later). Previous experiments suggest that the standard deviation of \(Y\) under a standard diet is approximately \(80 \mathrm{~g} .^{23}\) Using this as a guess of \(\sigma\), determine how many turkeys you should have in a treatment group, if you want the standard error of the group mean to be no more than (a) \(20 \mathrm{~g}\) (b) \(15 \mathrm{~g}\)

compute the standard error of \(\left(\bar{Y}_{1}-\bar{Y}_{2}\right)\) for the following data: $$ \begin{array}{|lcc|} \hline & \text { Sample 1 } & \text { Sample 2 } \\ \hline n & 10 & 10 \\ \bar{y} & 125 & 217 \\ s & 44.2 & 28.7 \\ \hline \end{array} $$

In a field study of mating behavior in the Mormon cricket (Anabrus simplex), a biologist noted that some females mated successfully while others were rejected by the males before coupling was complete. The question arose whether some aspect of body size might play a role in mating success. The accompanying table summarizes measurements of head width \((\mathrm{mm})\) in the two groups of females. \(^{45}\) (a) Construct a \(95 \%\) confidence interval for the difference in population means. [Note: Formula (6.7.1) yields 35.7 degrees of freedom for these data. (b) Interpret the confidence interval from part (a) in the context of this setting. (c) Using your interval computed in (a) to support your answer, is there strong evidence that the population mean head width is indeed larger for successful maters than unsuccessful maters? $$ \begin{array}{|lcc|} \hline & \text { Successful } & \text { Unsuccessful } \\ \hline n & 22 & 17 \\ \bar{y} & 8.498 & 8.440 \\ s & 0.283 & 0.262 \\ \hline \end{array} $$

Example 6.6 .3 reports measurements of pain for children who have had their tonsils removed. Another variable measured in that experiment was the number of doses of Tylenol taken by the children in the two groups. Those data are $$\begin{array}{|lcc|}\hline & {\text { Type of surgery }} \\\\{ 2 - 3 } & \text { Conventional } & \text { Coblation } \\\\\hline n & 49 & 52 \\\\\bar{y} & 3.0 & 2.3 \\\\\text { SD } & 2.4 & 2.0 \\\\\hline\end{array}$$ Compute the standard error of \(\left(\bar{Y}_{1}-\bar{Y}_{2}\right)\).

Prothrombin time is a measure of the clotting ability of blood. For 10 rats treated with an antibiotic and 10 control rats, the prothrombin times (in seconds) were reported as follows \(^{43}\) : $$ \begin{array}{|lcc|} \hline & \text { Antibiotic } & \text { Control } \\ \hline n & 10 & 10 \\ \bar{y} & 25 & 23 \\ s & 10 & 8 \\ \hline \end{array} $$ (a) Construct a \(90 \%\) confidence interval for the difference in population means. (Assume that the two populations from which the data came are normally distributed.) [Note: Formula (6.7.1) yields 17.2 degrees of freedom for these data.] (b) Why is it important that we assume that the two populations are normally distributed in part (a)? (c) Interpret the confidence interval from part (a) in the context of this setting.

See all solutions

Recommended explanations on Biology Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free