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points) A computer random number generator was used to generate 950 random digits (0,1 ,_1,9). The observed frequencies of the digits are given in the table below:
/100/97/79/89 102/78/101/90 133/
Test the claim that all the outcomes are equally likely using the significance level a = 0.01 , The expected frequency of each outcome is
Complete the table below to help calculate the test statistic. 0-E (O-E2
81
100
97
79
89
5/102
78

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03:02

A random number generator is supposed to produce the digits 0 through 9 with equal probability. A sample of 200 digits was generated, with the following frequency for each of the digits:Digit: 0 1 2 3 4 5 6 7 8 9Frequency: 21 17 20 18 25 16 28 19 22 14

At the 5% significance level, do these data provide evidence that the random number generator is not working properly? Explain.

07:19

( use excel or STATDISK). Generate 5000 digits and record theresults in the accompanying table. Use a 0.05 significance level totest the claim that the sample digits come from a population with auniform distribution ( so that all digits are equally likely.) Doesthe random number generator appear to be working as it shouldDigit0123456789Frequency

02:50

03:28

A computer scientist has developed an algorithm for generating pseudorandom integers over an interval of 0-9. He generates 1000 pseudorandom digits. Is there evidence that the random number generator is working correctly? The observed frequency is given below.

Random Number Observed Frequency0 941 932 1123 1014 1045 956 1007 998 1089 94

Answer the following questions based on the above information.

Q8) Write the test hypothesis.Q9) Perform a chi-square based goodness of fit test.Q10) Determine the p-value of the test.

Transcript

Hello students, in this question we have given the observed frequency in the table and we have to calculate the column for observed frequency minus expected and observed frequency minus expected bracket square divided by the expected frequencies. So for that first we need to compute the value of E that is the expected frequency and the expected frequency that is E is equal to capital N divided by K where capital N is the total of the observed frequency and that is 950 divided by K is the total number of categories that is 10 so this is equal to 95 and now by using this value of E we can calculate O minus E so this is 81 minus 95 that is equal to minus 14 then this is 5 2 minus 16 minus 6 7 minus 17 6 minus 5 and 38 so this is the column for the O minus E then O minus E bracket square divided by E is minus 14 square divided by the expected frequency that…

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