A walk-in medical clinic believes that arrivals are uniformly distributed over weekdays (Monday through Friday). It has collected the following data based on a random sample of 100 days. Frequency Mon25 Tue22 Wed19 Thu18 Fri16 Total100 Assuming that a goodness-of-fit test is to be conducted using a 0.10 level of significance, the critical value is:

The degrees of freedom are given by;

The significance level is 0.1 so then the critical value would be given by:

If the calculated value is higher than this value we can reject the null hypothesis that the arrivals are uniformly distributed over weekdays

Step-by-step explanation:

For this case we have the following observed values:

Mon 25 Tue 22 Wed 19 Thu 18 Fri 16 Total 100

For this case the expected values for each day are assumed:

The statsitic would be given by:

Where O represent the observed values and E the expected values

The degrees of freedom are given by;

The significance level is 0.1 so then the critical value would be given by:

If the calculated value is higher than this value we can reject the null hypothesis that the arrivals are uniformly distributed over weekdays

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step-by-step explanation:

answer:

step-by-step explanation:

here, the total dishes = 10

dishes having meat = 3 ( chicken, meatloaf and hamburger )

thus, the probability of having meat on the chosen dishes = number of dishes having meat / total dishes

=

therefore, the probability of not having meat = 1- the probability of having meat on the chosen dishes

=

=

=

⇒ option b is correct.