The wolfpack has one 11 games and lost 9 games this season. what percent of its games did the wolfpack win?

1 answer:

4 0

In fractions they had won 11/20 games. Then if you convert the fraction to a percent you would get 55% after multipling by 5/5 to get the denominator of 100.

The answer would be 55%.

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How many different shades of paint did the students make

Setler79 [48]

The correct answer should be nine

3 0

2 years ago

In a survey, the planning value for the population proportion is . How large a sample should be taken to provide a confidence in

tatuchka [14]

Answer:

n=350

Step-by-step explanation:

Notation and definitions

$n$ random sample taken (variable of interest)

$\hat p=0.35$ estimated proportion (value assumed)

$p$ true population proportion

Confidence =0.95 or 95% (value assumed)

Me=0.05 (value assumed)

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

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

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: