Question

	3. At Phil & Jims steak Shop in Parkside [best cheese steak on the planet] they sell about 750 cheese steak sandwiches per day.
	They expect that the mix of sandwiches will be 360 with fried onions only, 165 pizza steask, 120 mushroom steaks & 105 with no cheese.
	Last Wednesday they sold 370 with fried onions, 175 pizza steaks, 110 mushroom steaks & 95 with no cheese.
	At a 95% level of confidence, did their observations fit with their expectations?
	Ho:	fofits fe   (observed frequencies are consistent with those expected)
	H1:	fo doe not fit fe   (observed frequencies are not consistent with those expected)
		χ2 =
		χ2.05, 3 =
		Ho is		accepted/rejected
	Conclusion:

Answer 1

The null and alternative hypothesis is given as:

f₀ fits f_e , i.e., the observed frequencies(f₀) are consistent with those of expected frequencies(f_e).

f₀ does not fits f_e , i.e, the observed frequencies(f₀) are not consistent with those of expected frequencies(f_e).

Observed and Expected frequencies are given in the question as:

	fried onions	pizza steaks	mushroom steaks	No chesse	Total
Observed frequency	360	165	120	105	750
Expected frequency	370	175	110	95	750

The formula for the test-statistic is:      with degrees of freedom, df=k-1

The test-statistic is calculated as

Now in the question the confidence level is given as 95% which means the significance level is

The critical value with is:

Since,

Conclusion:

Since the test-statistic is less than the critical value, which means we fail to reject(FTR) null hypothesis H₀ . In other words, at , the data does not provide enough evidence to support alternative hypothesis . So we conclude that the observed and expected frequencies for the sandwiches per day are not different.