00 Chips Trial, claiming at every 18 bag of its tained at least 1000 chocolate c
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00 Chips Trial, claiming at every 18 bag of its tained at least 1000 chocolate chips 3tu hased random bags of iferent stores. ts p 079 Check the conditions for inference. Check the data for The sampl of the population. hat the data O B. The sample lass than 1% of the population sn nnA ranRssume that the data ara indapansant. that they ot independent. ren from na om mole, 80 one sume that they sre Construct of the the correct graph bclow. the trial ram Chocol correct 8wat bekrw. The hielegram rcughly unimodsl and aaymmetric, 8oone ca sume that the ts come pulsion that follows -Student's med B. The histog that follows orma model ly bi al and symrmotric, with ly bi dal and that the popula ion that follows -Student's ly unimadalanc symmatic, wit b) Create a 96% for the averag number of chips chipa (Round to decima place ceded ths evi y's claim? Choose the comect say about A. Beca idence inlet say that th ber of chips tan 1000 the N model predi nt of of chiR PRr hag is likely ryha9 st 1000 chips entirely above 1000, the ber of per ag is likely CCO. H the Normal model predicts that a small amount of vidual bags will D. Because the ccnfidence duces 1000, there is rel erough evidence that the m mbe of chip ie mcre 1000. Thia m inst every ba not cr tain 81 leaa chips mber of chil fewer than fewer th 00 chi 00 chips. of the data parts through below.Explanation / Answer
a.
Assumptions for inference:
The 10% condition states that sample sizes should be no more than 10% of thepopulation
B is right
b.
C is the right grph.
Look for where is max freq: it' near 1300 ( that' the mode - high bars should be here) - B is wrong
The expremes are above 1400, so A is wrong.
c.
D is the right description of histogram above
d. Z =2.055 for 98% percentile
So, Xbar +/- Z*Sigma/sqrt(n)
= 1258 +/- 2.055*117.51/sqrt(10)
=1181.64 to 1334.46, which doesn't include 1000
Since limits are above 1000, the range doesn't contain 1000.
B is the right interpretation.
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