/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 A penny was spun on a hard, flat... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A penny was spun on a hard, flat surface 50 times, and the result was 15 heads and 35 tails. Using a chisquare test for goodness of fit, test the hypothesis that the coin is biased, using a \(0.05\) level of significance.

Short Answer

Expert verified
The chi-square test for goodness of fit reveals that the coin is indeed biased because the calculated chi-square value (8) is greater than the critical chi-square value (3.84), at the 0.05 level of significance.

Step by step solution

01

Identify Observed and Expected Frequencies

The observed frequencies are given as 15 (Heads) and 35 (Tails). If the coin was unbiased, we would expect to see heads and tails equally, i.e., the expected frequencies would be 25 (50/2) each for Heads and Tails.
02

Apply Chi-square Goodness of Fit Formula

The Chi-square goodness of fit statistic is calculated as \(\chi^2 = \sum \frac{(O_i - E_i)^2}{ E_i }\) where \(O_i\) represents the observed frequency, and \(E_i\) the expected frequency. Substituting observed and expected values of Heads and Tails in the formula, it turns out to \(\chi^2 = (\frac{(15-25)^2}{25}) + (\frac{(35-25)^2}{25})\)
03

Calculate Chi-square Value

Calculate the chi-square value from the expression obtained in step 2, to get \(\chi^2 = 4+4 = 8\)
04

Compare with Chi-Square Distribution

Our calculated chi-square value of 8 needs to be tested against a chi-square distribution with 1 degree of freedom (number of outcomes - 1). From chi-square distribution tables, the critical value at 0.05 significance level and 1 degree of freedom is approximately 3.84.
05

Make a Decision

Since our calculated chi-square value (8) is greater than the critical chi-square value (3.84), we reject the null hypothesis. This means the number of heads and tails observed is significantly different from what would be expected if the coin was unbiased.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Rats had a choice of freeing another rat or eating chocolate by themselves. Most of the rats freed the other rat and then shared the chocolate with it. The table shows the data concerning the gender of the rat in control. $$\begin{array}{lcc} & \text { Male } & \text { Female } \\\\\hline \text { Freed Rat } & 17 & 6 \\\\\text { Did not } & 7 & 0 \\ \hline\end{array}$$ a. Can a chi-square test for homogeneity or independence be performed with this data set? Why or why not? b. Determine whether the sex of a rat influences whether or not it frees another rat using a significance level of \(0.05\).

The 2018 Pew Research poll in exercise \(10.43\) also reported responses by political party. Survey results found \(45 \%\) of Republicans and \(69 \%\) of Democrats supported marijuana legalization. a. Use these results to fill in the following two-way table with the counts in each category. Assume the sample size for each group was \(200 .\) b. Test the hypothesis that support of marijuana legalization is independent of political party for these two groups using a significance level of \(0.05 .\) c. Does this suggest that these political parties differ significantly in their support of marijuana legalization?

There is a theory that relative finger length depends on testosterone level. The table shows a summary of the outcomes of an observational study that one of the authors carried out to determine whether men or women were more likely to have a ring finger that appeared longer than their index finger. Identify both of the variables, and state whether they are numerical or categorical. If numerical, state whether they are continuous or discrete. $$\begin{array}{|lcc|}\hline & \text { Men } & \text { Women } \\\\\hline \text { Ring Finger Longer } & 23 & 13 \\\\\hline \text { Ring Finger Not Longer } & 4 & 14 \\\\\hline\end{array}$$

A 2018 Gallup poll asked college graduates if they agreed that the courses they took in college were relevant to their work and daily lives. The respondents were also classified by their field of study. If we wanted to test whether there was an association between response to the question and the field of study of the respondent, should we do a test of independence or homogeneity?

When playing Dreidel, (see photo) you sit in a circle with friends or relatives and take turns spinning a wobbly top (the dreidel). In the center of the circle is a pot of several foil-wrapped chocolate coins. If the four-sided top lands on the Hebrew letter gimmel, you take the whole pot and everyone needs to contribute to the pot again. If it lands on hey, you take half the pot. If it lands on \(n u n\), nothing happens. If it lands on shin, you put a coin in. Then the next player takes a turn. Each of the four outcomes is believed to be equally likely. One of the author's families got the following outcomes while playing with a wooden dreidel during Hanukah. Determine whether the outcomes allow us to conclude that the dreidel is biased (the four outcomes are not equally likely). Use a significance level of \(0.05 .\) $$\begin{array}{cccc}\text { gimmel } & \text { hey } & \text { nun } & \text { shin } \\\5 & 1 & 7 & 27\end{array}$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.