Tara earns $8.50 an hour (after taxes) at the pizza place. She is scheduled to work four hours this afternoon. However, her frie

Answer:

For the first year, the total amount comes to $266, and with an hourly wage of $5.50, you can express this as the equation 5.50x=266

Step-by-step explanation:

Answer:

Step-by-step explanation:

The goal here is to identify whether the provided data set adheres to a normal distribution, and consequently, if a t-test is suitable for application.

To address this, we will construct a box plot using MINITAB software

The guidelines for this are as follows:

Input your data into the worksheet

Then navigate to Graph→ Boxplot→ EDA→ Boxplot

Under One Y, choose simple, and click OK

In Variables, enter measurements. Then click OK

Thus, for the first dataset (a) 213.03 212.95 213.04 213.00 212.99 213.01 221.03 213.05 as illustrated in the attached file in word document for the boxplot below;

We would observe an outlier in the data set which suggests that it does not conform to the normal distribution assumption; hence, the use of the t-distribution is not advised.

(b) Referring to the attached document below; we will notice that the boxplot does not present any outliers. Furthermore, it is evident that the distribution is right skewed. Thus, this dataset fails the necessary assumptions, making the t-distribution inappropriate for this data.

(c)

In the document provided below; we will find the boxplot without any outliers. Additionally, it is clear that the distribution exhibits a left skew. Consequently, the data does not meet the required assumptions, resulting in the t-distribution being unsuitable for this dataset.

Answer:

Step-by-step explanation:

It is known that the mean and standard deviation of the sampling distribution of the sample proportion() are represented as follows:-

, where p= Population proportion and n = sample size.

Let p denote the proportion of blue chips.

According to the information provided, we have

p= 0.275

n= 5

Thus, the mean and standard deviation of the sampling distribution of the sample proportion of blue chips for samples of size 5 will be:

Therefore, you will have the mean and standard deviation for the sample proportion of blue chips for samples of size 5: