a statistician just bought a new car with a claimed average gas mileage of 39 miles per gallon. the first 10…

a statistician just bought a new car with a claimed average gas mileage of 39 miles per gallon. the first 10 times he fills up the gas tank he calculates the average gas mileage (in miles per gallon, or mpg) and records it; the values are below. 37, 38.1, 36.8, 37.9, 35.1, 36.8, 38.9, 39.8, 37.9, 37.4. all of the following questions will refer to this data - set. you should enter it into rstudio and confirm that youve entered it correctly.\nwhat is the population being studied?\n○ all cars of the same model as this one\n○ all statisticians who own cars\n○ gas mileage\n○ all times that this car is ever filled with gas\nwhat is the variable being studied?\n○ number of times the gas tank is filled\n○ average miles per gallon\n○ miles per gallon\n○ gallons of gas\nwhat is the mean of the data - set? round to two decimal places. \nwhat is the standard deviation of the data - set? round to four decimal places. \nwhat is the shape of the data - set?\n○ right - skewed.\n○ left - skewed.\n○ roughly symmetric.\nare there any outliers in the data set (hint: make a boxplot in rstudio).\n○ yes\n○ no\nthe statistician collected the data to see if his car has an average gas mileage of 39 or not. perform a hypothesis test to determine whether there is evidence for this, making sure to include all steps:\nstep 1. state hypotheses.\n○ h0:μ = 39 and ha:μ≠39\n○ h0:μ = 39 and ha:μ<39\n○ h0:μ = 37.57 and ha:μ>37.57\n○ h0:μ = 37.57 and ha:μ≠37.57
Answer
Explanation:
Step1: Calculate the mean
The mean $\bar{x}$ of a data - set $x_1,x_2,\cdots,x_n$ is given by $\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}$. Here, $n = 10$ and the data - set is $37,38.1,36.8,37.9,35.1,36.8,39.9,39.8,37.9,37.4$. $\sum_{i=1}^{10}x_i=37 + 38.1+36.8+37.9+35.1+36.8+39.9+39.8+37.9+37.4 = 376.7$ $\bar{x}=\frac{376.7}{10}=37.67$
Step2: Calculate the standard deviation
The formula for the sample standard deviation $s$ is $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}$. First, calculate $(x_i-\bar{x})^2$ for each $i$: $(37 - 37.67)^2=(-0.67)^2 = 0.4489$ $(38.1-37.67)^2=(0.43)^2 = 0.1849$ $(36.8 - 37.67)^2=(-0.87)^2 = 0.7569$ $(37.9-37.67)^2=(0.23)^2 = 0.0529$ $(35.1-37.67)^2=(-2.57)^2 = 6.6049$ $(36.8 - 37.67)^2=(-0.87)^2 = 0.7569$ $(39.9-37.67)^2=(2.23)^2 = 4.9729$ $(39.8-37.67)^2=(2.13)^2 = 4.5369$ $(37.9-37.67)^2=(0.23)^2 = 0.0529$ $(37.4-37.67)^2=(-0.27)^2 = 0.0729$ $\sum_{i = 1}^{10}(x_i-\bar{x})^2=0.4489+0.1849+0.7569+0.0529+6.6049+0.7569+4.9729+4.5369+0.0529+0.0729 = 18.431$ $s=\sqrt{\frac{18.431}{9}}\approx1.4361$
Step3: Determine the shape
By looking at the data values and potentially creating a histogram or box - plot, we can see that the data is roughly symmetric.
Step4: Check for outliers
By creating a box - plot in RStudio (or using the 1.5 * IQR rule), we can find that there are no outliers.
Step5: Hypothesis testing
The null hypothesis $H_0:\mu = 39$ and the alternative hypothesis $H_a:\mu\neq39$ The test statistic for a one - sample t - test is $t=\frac{\bar{x}-\mu}{s/\sqrt{n}}$, where $\bar{x}=37.67$, $\mu = 39$, $s\approx1.4361$, and $n = 10$. $t=\frac{37.67 - 39}{1.4361/\sqrt{10}}\approx\frac{- 1.33}{0.454}\approx - 2.93$ The degrees of freedom is $n - 1=9$. Using a t - distribution table or R function (e.g., 2*pt(-2.93,9)), the p - value is approximately $0.018$. Since the p - value is less than the common significance level of 0.05, we reject the null hypothesis.
Answer:
- Mean: 37.67
- Standard deviation: 1.4361
- Shape: Roughly symmetric
- Outliers: No
- Hypothesis test: Reject $H_0:\mu = 39$ (p - value $\approx0.018$)