Complete the table. HINT: its not 10 or 21

A coin is tossed 71 times and 34 heads are observed. Is the coin fair? Use a 98% level confidence interval to base your inferenc

Fofino [41]

Answer:

a) $\hat p=\frac{34}{71}=0.48$ estimated proportion of heads observed.

b) $Se= \sqrt{\frac{0.479(1-0.479)}{71}}=0.0593$

c) $z_{\alpha/2}=2.33$

d) $0.479 - 2.33 \sqrt{\frac{0.479(1-0.479)}{71}}=0.341$

e) $0.479 + 2.33 \sqrt{\frac{0.479(1-0.479)}{71}}=0.617$

f) In order to be fair the coin needs to have included the 0.5 on the interval, and for this case it's included, so we can say that at 2% of significance we can say that the coin is fair.

Step-by-step explanation:

Notation and definitions

$X=34$ number of heads observed.

$n=71$ random sample taken

$\hat p=\frac{34}{71}=0.479$ estimated proportion of heads observed.

$p$ true population proportion of heads observed.

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

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

The point estimator for the population proportion of heads is: Answer with two decimal precision.

$\hat p=\frac{34}{71}=0.48$ estimated proportion of heads observed.

The standard error in this estimate is: Answer with four decimal precision.

The confidence interval would be given by this formula

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

And the standard error is given by:

$Se= \sqrt{\frac{\hat p(1-\hat p)}{n}}$

$Se= \sqrt{\frac{0.479(1-0.479)}{71}}=0.0593$

The correct table value for the 98% level confidence interval is: Answer with two decimal precision

For the 98% confidence interval the value of $\alpha=1-0.98=0.02$ and $\alpha/2=0.01$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.33$

The lower end point of the CI is: Answer with three decimal precision.

And replacing into the confidence interval formula we got:

$0.479 - 2.33 \sqrt{\frac{0.479(1-0.479)}{71}}=0.341$

The upper end point of the CI is: Answer with three decimal precision.

$0.479 + 2.33 \sqrt{\frac{0.479(1-0.479)}{71}}=0.617$

And the 90% confidence interval would be given (0.341;0.617).

Based on the confidence interval, it is plausible that the coin is fair.

In order to be fair the coin needs to have included the 0.5 on the interval, and for this case it's included, so we can say that at 2% of significance we can say that the coin is fair.

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