Some students checked 6 bags of Doritos marked with a net weight of 28.3 grams. They carefully weighed the contents of each bag, recording the following weights (in grams): 29.2, 28.5, 28.7, 28.9, 29.1, 29.5.

Why might finding a bootstrap confidence interval not be a good idea for these data?

Some students checked 6 bags of Doritos marked with a net weight of 28.3 grams. They carefully weighed the contents of each bag, recording the following weights (in grams): 29.2, 28.5, 28.7, 28.9, 29.1, 29.5.

Create a 95% confidence interval for the mean weight of such bags of chips.

Some students checked 6 bags of Doritos marked with a net weight of 28.3 grams. They carefully weighed the contents of each bag, recording the following weights (in grams): 29.2, 28.5, 28.7, 28.9, 29.1, 29.5 .

Find the mean and standard deviation of the weights.

Some students checked 6 bags of Doritos marked with a net weight of 28.3 grams. They carefully weighed the contents of each bag, recording the following weights (in grams): 29.2, 28.5, 28.7, 28.9, 29.1, 29.5 .

Test the hypothesis that the net weight is as claimed.

Software analysis of the salaries of a random sample of 288 Nevada teachers produced the confidence interval shown below. Which conclusion is correct? What’s wrong with the others?

$$\text { t-Inter val for } \mu : \left. \begin{array} { c c } { \text {with } 90.00 \% \text { Confidence, } } \\ { } { 38944 < \mu [ \text { TchPay]} < 42893 } \end{array} \right.$$

a) If we took many random samples of 288 Nevada teachers, about 9 out of 10 of them would produce this confidence interval. b) If we took many random samples of Nevada teachers, about 9 out of 10 of them would produce a confidence interval that contained the mean salary of all Nevada teachers. c) About 9 out of 10 Nevada teachers earn between $38,944 and$42,893. d) About 9 out of 10 of the teachers surveyed earn between $38,944 and$42,893. e) We are 90% confident that the average teacher salary in the United States is between $38,944 and$42,893.

Software analysis of the salaries of a random sample of 288 Nevada teachers produced the confidence interval shown below. Which conclusion is correct? What's wrong with the others? t-Interval for

$$\mu$$

: with 90.00 % Confidence, 38944 <

$$\mu$$

(TchPay) < 42893. If we took many random samples of Nevada teachers, about 9 out of 10 of them would produce a confidence interval that contained the mean salary of all Nevada teachers.

Software analysis of the salaries of a random sample of 288 Nevada teachers produced the confidence interval shown below. Which conclusion is correct? What's wrong with the others?

$$t \text {-Interval for } \mu \text { : } \begin{aligned} & \text { with } 90.00 \% \text { Confidence, } \\ & 43454<\mu(\text { TchPay })<45398 \end{aligned}$$

If we took many random samples of 288 Nevada teachers, about 9 out of 10 of them would produce this confidence interval.

Software analysis of the salaries of a random sample of 288 Nevada teachers produced the confidence interval shown below. Which conclusion is correct? What's wrong with the others? t-Interval for

$$\mu$$

: with 90.00 % Confidence, 38944 <

$$\mu$$

(TchPay) < 42893. About 9 out of 10 of the teachers surveyed earn between $38,944 and$42,893.

Software analysis of the salaries of a random sample of 288 Nevada teachers produced the confidence interval shown below. Which conclusion is correct? What's wrong with the others? t-Interval for

$$\mu$$

: with 90.00 % Confidence, 38944 <

$$\mu$$

(TchPay) < 42893. About 9 out of 10 Nevada teachers earn between $38,944 and$42,893..

Software analysis of the salaries of a random sample of 288 Nevada teachers produced the confidence interval shown below. Which conclusion is correct? What's wrong with the others? t-lnterval for

$$\mu$$

: with 90.00% Confidence,

$$38944 < \mu ( \text { TchPay } ) < 42893$$

We are 90% confident that the average teacher salary in the United States is between $38,944 and$42,893.

During routine screening, a doctor notices that 22% of her adult patients show higher than normal levels of glucose in their blood-a possible warning signal for diabetes. Hearing this, some medical researchers decide to conduct a large-scale study, hoping to estimate the proportion to within 4% with 98% confidence. How many randomly selected adults must they test.

During routine screening, a doctor notices that 22% of her adult patients show higher than normal levels of glucose in their blood—a possible warning signal for diabetes. Hearing this, some medical researchers decide to conduct a large-scale study, hoping to estimate the proportion to within 4% with 98% confidence. How many randomly selected adults must they test?

Transcription is the process of RNA synthesis. Translation is the cytoplasmic process of protein synthesis. Both transcription and translation can be divided into three stages: initiation, elongation, and termination. Draw a table comparing the different processes at each stage of transcription and translation.

In a study comparing speech-to-text transcription services, researchers estimated the accuracy rate of YouTube's auto-transcription service to be 72 % (28% of words were incorrectly transcribed), with a 95 % confidence interval of (67.8 %, 75.5 %).

"Explain what " 95 % confidence" means.