Question

The following data represent the pH of rain for a random sample of 12 rain dates. A normal probability plot suggests the data could come from a population that is normally distributed.
5.20, 5.02, 4.87, 5.72, 4.57, 4.76, 4.99, 4.74, 4.56, 4.80, 5.19, 4.68
1) Determine a point estimate for the population mean.
2) Construct and Interpret a 95% confidence interval for the mean pH of rainwater.
a) if repeated samoles are taken, 95% of them will have a sample pH of rain water between [ ] & [ ].
b) there is a 95% chance that the true mean pH of rain water is between [ ] & [ ].
c) there is 95% confidence that the population mean pH of rain water is between [ ] & [ ].
3) Construct and interpret a 99% confidence interval for the mean pH of rainwater.
a) there is 99% confidence that the population mean pH of rain water is between [ ] & [ ].
b) there is a 99% chance that the true mean pH of rain water is between [ ] & [ ].
c) if repeated samoles are taken, 99% of them will have a sample pH of rain water between [ ] & [ ].
4) What happens to the interval as the level of confidence is changed? Explain why is a logical result.
As the level of confidence increases l, the width of the interval_____this makes sense since the_____,______.

Answer 1

Answer:

(1) The point estimate for the population mean is 4.925.

(2) Therefore, a 95% confidence interval for the population mean pH of rainwater is [4.715, 5.135] .

(3) Therefore, a 99% confidence interval for the population mean pH of rainwater is [4.629, 5.221] .

(4) As the level of confidence increases, the width of the interval increases.

Step-by-step explanation:

We are given that the following data represent the pH of rain for a random sample of 12 rain dates.

X = 5.20, 5.02, 4.87, 5.72, 4.57, 4.76, 4.99, 4.74, 4.56, 4.80, 5.19, 4.68.

(1) The point estimate for the population mean is given by;

Point estimate, $\bar X$ = $\frac{\sum X}{n}$

                            = $\frac{5.20+5.02+ 4.87+5.72+ 4.57+ 4.76+4.99+ 4.74+ 4.56+ 4.80+5.19+ 4.68}{12}$

                            = $\frac{59.1}{12}$  = 4.925

(2) Let $\mu$ = mean pH of rainwater

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                             P.Q.  =  $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$  ~  $t_n_-_1$

where, $\bar X$ = sample mean = 4.925

             s = sample standard deviation = 0.33

            n = sample of rain dates = 12

            $\mu$ = population mean pH of rainwater

Here for constructing a 95% confidence interval we have used a One-sample t-test statistics as we don't know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-2.201 < $t_1_1$ < 2.201) = 0.95  {As the critical value of t at 11 degrees of

                                               freedom are -2.201 & 2.201 with P = 2.5%}  

P(-2.201 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.201) = 0.95  

P( $-2.201 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.201 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.201 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.201 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-2.201 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.201 \times {\frac{s}{\sqrt{n} } }$ ]

                                      = [ $4.925-2.201 \times {\frac{0.33}{\sqrt{12} } }$ , $4.925+2.201 \times {\frac{0.33}{\sqrt{12} } }$ ]  

                                     = [4.715, 5.135]

Therefore, a 95% confidence interval for the population mean pH of rainwater is [4.715, 5.135] .

The interpretation of the above confidence interval is that we are 95% confident that the population mean pH of rainwater is between 4.715 & 5.135.

(3) Now, 99% confidence interval for the population mean, $\mu$ is ;

P(-3.106 < $t_1_1$ < 3.106) = 0.99  {As the critical value of t at 11 degrees of

                                               freedom are -3.106 & 3.106 with P = 0.5%}  

P(-3.106 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 3.106) = 0.99

P( $-3.106 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $3.106 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

P( $\bar X-3.106 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+3.106 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

99% confidence interval for $\mu$ = [ $\bar X-3.106 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+3.106 \times {\frac{s}{\sqrt{n} } }$ ]

                                     = [ $4.925-3.106 \times {\frac{0.33}{\sqrt{12} } }$ , $4.925+3.106 \times {\frac{0.33}{\sqrt{12} } }$ ]  

                                     = [4.629, 5.221]

Therefore, a 99% confidence interval for the population mean pH of rainwater is [4.629, 5.221] .

The interpretation of the above confidence interval is that we are 99% confident that the population mean pH of rainwater is between 4.629 & 5.221.

(4) As the level of confidence increases, the width of the interval increases as we can see above that the 99% confidence interval is wider as compared to the 95% confidence interval.