A die that is fair will have each face of the die come up one-sixth of the time in the long run. The population for die throwing contains the results of throwing the die an infinite number of times. For this problem, the parameter of interest is p is the probability of rolling a 4, which is the proportion of all tosses in the population that have a 4 showing on the upturned face of the die.
Suppose that you have a die that may or may not be fair. On a rainy day, you toss the die 1500 times and record the outcome of each toss. After many hours and a tired arm, you found that you had rolled 282 fours. Using the following steps, conduct a hypothesis test to determine the strength of the evidence that the die is not fair.
Ho: p=1/6
Ha: p not equal to 1/6
What is the z-statistic for this test?
SE of proportion = sq rt of (pq/n)
when p = proprotion of 6s
and q = proportion of other values
I hope this helps.
A card is picked at random from a pack of 20 cards numbered 1-20. given that the cards shows an even number, find the probability that it is a multiple of 4
To determine the z-statistic for this hypothesis test, we need to calculate the test statistic based on the sample data and the null hypothesis.
The formula for the z-statistic in this case is:
z = (p̂ - p) / √(p(1 - p) / n)
Where:
- p̂ is the sample proportion (number of successes / total observations).
- p is the hypothesized proportion under the null hypothesis.
- n is the sample size.
Given the information provided:
- The sample proportion (p̂) is the number of fours rolled (282) divided by the total number of tosses (1500): p̂ = 282 / 1500.
- The hypothesized proportion (p) is 1/6, as stated in the null hypothesis.
- The sample size (n) is 1500.
Now we can substitute these values into the formula to calculate the z-statistic:
z = [(282 / 1500) - (1/6)] / √[(1/6)(1 - 1/6) / 1500]
Calculating this expression will give you the value of the z-statistic, which can be used to assess the strength of the evidence against the null hypothesis.