Question

A​ renters' organization claims that the median number of rooms in​ renter-occupied units is

five.You randomly select 40​renter-occupied units and obtain the results shown below. At

alphα=0.02, complete parts​ (a) through​ (f).

Unit Size	Number of units
Fewer than 5 rooms	12
5 rooms	11
More than 5 rooms	17

a. The claim is

b. What are Ho and Ha?

c. crticial value

d. test stat

e.Decide whether to reject or fail to reject the null hypothesis.

f.  At the

2 %2%

significance​ level,

▼

there is not

there is

enough evidence to reject the​ organization's claim​ "The median number of rooms in​ renter-occupied units is

fivefive​."

Answer 1

(a) claim is the median number of rooms in​ renter-occupied units is 5

(b)

alternative hypothesis, H0: the median number of rooms is 8.

alternative hypothesis is Ha: the median number if rooms is not 8

To run the test:

Step 1: Assign a value of 0 to those with rooms below your hypothesized median, and 1, to those above.

so we get 17

Step 2: Find the obtained frequency (of). There are 12 below the median (the median is 5), and 17 above.
of = 17 and 12

Step 3: Find the expected frequency (ef). The total number of data point is 40 if 5 was a true median, we’d expect to have 20 rooms having below and 20 above
ef = 20

Step 4: Calculate the chi-square:

Χ2 = Σ [(of – ef)2/ef].

This is:
(17– 5)²/5+ (12 – 5)²/5

or 38.6

Step 5: Find the degrees of freedom. There are two observed frequencies (equal to two cells in a contingency table), so there is just one degree of freedom.

Step 6: Use a Chi-squared table to find the critical chi-square value
for 1 degree of freedom and an alpha level of 0.02 (α = 0.02). This equals 57.96

Since our Χ2 value is less than the critical Χ2 value,we fail to reject the null hypothesis (i.e. we can keep it) —that the median is 5. If the Χ2 value we calculated from our data had been more than the critical Χ2 value, we would be able to reject the null hypothesis and postulated median age of 5

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