Question

In a study conducted in the United Kingdom about sleeping positions, 1000 adults in the UK were asked their starting position when they fall asleep at night. The most common answer was the fetal position (on the side, with legs pulled up), with of the participants saying they start in this position. Use a normal distribution to find a 95% confidence interval for the proportion of all UK adults who start sleep in this position. Use the fact that the standard error of the estimate is 0.016.

Answer 1

Answer:

The 95% confidence interval would be given by (0.380;0.441)

Step-by-step explanation:

In a study conducted in the United Kingdom about sleeping positions, 1000 adults in the UK were asked their starting position when they fall asleep at night. The most common answer was the fetal position (on the side, with legs pulled up), with 41% of the participants saying they start in this position. Use a normal distribution to find and interpret a 95% confidence interval for the proportion of all UK adults who start sleep in this position. Use the fact that the standard error of the estimate is 0.016

1) Notation and definitions

$X$ number of people who answer fetal position

$n=1000$ random sample taken

$\hat p=0.41$ estimated proportion of people who answer fetal position

$p$ true population proportion of people who answer fetal position

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

2) Confidence interval

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$. And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The confidence interval for the mean is given by the following formula:  

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.41 - 1.96\sqrt{\frac{0.41(1-0.41)}{1000}}=0.380$

$0.41 + 1.96\sqrt{\frac{0.41(1-0.41)}{1000}}=0.441$

The 95% confidence interval would be given by (0.380;0.441)

Answer 2

The confidence interval at the 95% level of confidence is (0.380 ; 0.441)

The confidence interval is defined thus :

• C. I = p ± Z*σ
• p = 41% = 0.41

Standard deviation :

• $\sqrt{\frac{p(1 - p)}{n}}$

Hence,

• σ = $\sqrt{\frac{0.41(0.59)}{1000}} = 0.0156$

• Z* at 95% = 1.96

Lower boundary = 0.41 - 1.96(0.0156) = 0.379

Upper boundary = 0.41 + 1.96(0.0156) = 0.440

Therefore, the confidence interval is (0.380; 0.441)

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