The sample mean \(\bar{x}\) can be found by finding the average of the population mean from the confidence interval.
\(\frac{65 +77}{2} = 71\).
\(\bar{x} = 71\)
To find the critical \(t^*\) value, I used the qt() function to find the value of \(t^*\) with .05% in each tail. \(t^* = 1.71\)
qt(p=.05, df=24, lower.tail = F)## [1] 1.710882\(71 \pm 1.71\times \frac{\sigma}{\sqrt25}\)
\(\sigma = 17.5\)
Margin of Error = \(t^* \times \frac{\sigma}{\sqrt{25}}\) = 5.99
Since I don’t know the sample size, I can’t calculate the degrees of freedom for the \(t\) value. I used the \(z\) table instead. With a 90% CI, I split the tails in two and find the z-value with a probability of 95%. \(z = 1.65\)
Margin of Error = \(z^* \times \frac{250}{\sqrt{n}}\)
Sample size = \((\frac{1.64 \times 250}{25})^2\)
Sample size = 269
z <- 1.65
size <- ((z*250)/25)^2
size## [1] 272.25From the z-table, the z-score would be 2.57. From looking at the equation above, to keep the margin of error the same the sample size would have to be larger.
z <- 2.58
size <- ((z*250)/25)^2
size## [1] 665.64From the boxplots we see that the average writing scores are higher than the average reading scores.
Since the reading and writing scores are from the same student, the scores are dependent.
\(H_o = 0\) There is no difference between reading and writing scores.
\(H_a \neq 0\) There is a difference between reading and writing scores.
The sample was random and was less than 10% of the population making it independent
The distribution of the data is normal and not skewed.
Test statistic:
\(T = \frac{-0.545 - 0}{8.887} = -.061\)
df = 199
p-value = .476
Since the p-value is so large, there is not enough evidence to say there is a convincing difference between the average scores.
pnorm(-.061)## [1] 0.4756796We may have made a Type II error meaning that there was a difference in the average test scores that we failed to detect.
No, because the p-value is so large the confidence interval would have to be quite low to have a chance to reject the null hypothesis.
\(H_o = 0\) :There is no significant difference between the average MPG
\(H_a \neq 0\) :There is a significant difference between the average MPG
m_auto <- 16.12
m_man <- 19.85
sd_auto <- 3.58
sd_man <- 4.51
n <- 26
df <- 25
diff_mean <- m_man - m_auto
se <- ((4.51^2)/26) + ((3.58^2)/26)
se <- sqrt(se)
print(paste0("Point estimate is: ", diff_mean))## [1] "Point estimate is: 3.73"print(paste0("Standard error is: ", round(se,2)))## [1] "Standard error is: 1.13"#95% confidence interval
t <- 2.06
x_high <- diff_mean + (t*se)
x_low <- diff_mean - (t*se)
print(paste0("We are 95% confident that the gas mileage in the manual is between ", round(x_low,2), "% and ", round(x_high,2), "% better than an automatic car."))## [1] "We are 95% confident that the gas mileage in the manual is between 1.4% and 6.06% better than an automatic car."t_score <- (diff_mean - 0)/se
p <- pnorm(-t_score)*2
print(paste0("p-value is: ", round(p,5)))## [1] "p-value is: 0.00096"print("Since the p-value is lower than the significant value of 0.05, there is a significant differenct between the average gas milage of the automatic and manual cars.")## [1] "Since the p-value is lower than the significant value of 0.05, there is a significant differenct between the average gas milage of the automatic and manual cars."\(H_o:\) Mean is equal across all groups
\(H_a:\) At least one mean is different
The observations are independent within and across groups
The data within each group are nearly normal
The variabliliy across the groups is about equal
\(F = \frac{MSG}{MSE}\)
\(MS_d = \frac{SS_d}{Df_d}\)
\(MS_r = \frac{SS_r}{Df_r}\)
\[ \begin{array}{ c | c | c | c | c | c } \hline & Df & Sum \ Sq & Mean \ Sq & F \ value & Pr(>F) \\ \hline degree & 4 & 2,006.16 & 501.54 & 0.0225 & 0.0682 \\ Residuals & 1167 & 267,382 & 229.12 \\ \hline Total & 1171 & 269,388.16 \end{array} \]
Since the p-value is larger than the F value, we reject H_a. The data didn’t provide significant evidence that the average hours worked differed between groups with different degrees.