This week's topics covered **tests for population proportion** and **population mean**, **statistical significance**, and
using a **confidence interval** to draw a conclusion about a two-tailed test.

### Question # 1 : The director of manufacturing at a cookie factory needs to determine whether a new machine is producing a particular type of cookie according to the manufacturer's specifications, which indicate that cookies should have a mean of 70 and standard deviation of 3.5 pounds. A sample pf 49 of cookies reveals a sample mean breaking strength of 69.1 pounds.

#### State the null and alternative hypothesis _____.

H_{0}: The machine is producing according to the manufacterer's specifications (μ ≥ 70). H_{a}: The machines is not producing according to the manufacturer's specifications (μ < 70).

#### Is there evidence that the machine is not meeting the manufacturer's specifications for average strength? Use a 0.05 level of significance _____.

> mu <- 70 #population mean > sd <- 3.5 #standard deviation > n <- 49 # sample size > xhat <- 69.1 #sample mean > alpha <- 0.05 > c <- 0.95 #critical value > test_stat <- ((xhat - mu) / (sd/sqrt(n))) > z <- qnorm(c + (alpha/2)) * -1 > cat("The test statistic is ", test_stat, " and the z value is ", z, ". We cannot reject the null hypothesis because the test statistic is greater than the z value statistic.") The test statistic is -1.8 and the z value is -1.959964 . We cannot reject the null hypothesis because the test statistic is greater than the z value statistic.

#### Compute the p value and interpret its meaning _____.

> p <- pnorm(test_stat) > p [1] 0.03593032 # The p value is < α (0.05), so we can reject the null hypothesis.

#### What would be your answer in (B) if the standard deviation were specified as 1.75 pounds?

> sd <- 1.75 > test_stat <- ((xhat - mu) / (sd/sqrt(n))) > cat("The test statistic is ", test_stat, " and the z value is ", z, ". We can reject the null hypothesis because the test statistic is lesser than the z value statistic.") The test statistic is -3.6 and the z value is -1.959964. We can reject the null hypothesis because the test statistic is lesser than the z value statistic.

#### What would be your answer in (B) if the sample mean were 69 pounds and the standard deviation is 3.5 pounds?

> xhat <- 69 > sd <- 3.5 > test_stat <- ((xhat - mu) / (sd/sqrt(n))) > cat("The test statistic is ", test_stat, " and the z value is ", z, ". We can reject the null hypothesis because the test statistic is lesser than the z value statistic.") The test statistic is -2 and the z value is -1.959964. We can reject the null hypothesis because the test statistic is lesser than the z value statistic.

### Question # 2 : If x̅ = 85, σ = standard deviation = 8, and n=64, set up 95% confidence interval estimate of the population mean μ.

> xhat <- 85 #sample mean > sd <- 8 #population standard deviation > n <- 64 #sample size > z <- 1.96 > me <- z*(sd/sqrt(n)) > lower <- xhat - me > upper <- xhat + me > cat("We can say with 95% confidence that the population mean is between (", lower, ", ", upper, ")") We can say with 95% confidence that the population mean is between ( 83.04 , 86.96 )

### Question # 3 : The accompanying data are: x= girls and y =boys. (goals, time spend on assignment)

#### Calculate the correlation coefficient for this data set _____.

> data <- read.csv(file = "m8.csv") > print(data) X Girl1 Girl2 Girl3 Boy1 Boy2 Boy3 1 Goals 4 5 6 4.0 5.0 6.0 2 Grades 49 50 69 46.1 54.2 67.7 3 Popular 24 36 38 26.9 31.6 39.5 4 Time spend on assignment 19 22 28 18.9 22.2 27.8 5 Total 92 108 135 95.9 113.0 141.0 > x <- c(data[["Girl1"]][5],data[["Girl2"]][5],data[["Girl3"]][5]) > y <- c(data[["Boy1"]][5],data[["Boy2"]][5],data[["Boy3"]][5]) > dframe <- data.frame(x,y) > corr <- cor(x,y) > cat("The correlation coefficient is: ", corr) The correlation coefficient is: 0.9999681

#### Pearson correlation coefficient _____.

> pearson <- cor(dframe, method="pearson") > cat("The Pearson correlation coefficient is: ", pearson) The Pearson correlation coefficient is: 1 0.9999681 0.9999681 1

#### Create plot of the correlation.

> plot(x, y, xlab="Girls", ylab="Boys", pch=21)